Explanation:
Okay, so I need to figure out how many of the same digits are found in the base 7 and base 8 representations of the decimal number 629. Let me start by recalling how to convert numbers between different bases.

First, I should convert 629 from base 10 to base 7 and then to base 8. Once I have both representations, I can compare the digits to see which ones are common. The question is asking for the count of digits that are the same in both representations.

Starting with base 7 conversion. To convert a decimal number to another base, I can use the division-remainder method. Let me try that.

So, for base 7:

Divide 629 by 7 and record the remainders.

First division: 629 ÷ 7. Let me calculate that. 7*89 is 623, right? 7*90 is 630, which is 1 more than 629. So 629 ÷ 7 is 89 with a remainder of 6. So the least significant digit (rightmost digit) is 6.

Now take the quotient, which is 89, and divide by 7 again. 7*12 is 84, so 89 ÷ 7 is 12 with a remainder of 5. So the next digit is 5.

Take the quotient 12 and divide by 7. 7*1 is 7, so 12 ÷ 7 is 1 with a remainder of 5. Next digit is 5.

Take the quotient 1 and divide by 7. That's 0 with a remainder of 1. So the next digit is 1.

Since the quotient is now 0, we stop. So writing the remainders from last to first, the base 7 representation is 1 5 5 6. So 1556 base 7. Let me check that.

Let me verify: 1*7^3 + 5*7^2 + 5*7^1 + 6*7^0.

Calculating each term:

1*343 = 343

5*49 = 245

5*7 = 35

6*1 = 6

Adding them up: 343 + 245 = 588; 588 + 35 = 623; 623 + 6 = 629. Perfect, that's correct. So base 7 is 1556.

Now moving on to base 8. Let me do the same process for base 8.

Convert 629 to base 8.

Divide 629 by 8. Let me calculate 8*78 = 624. So 629 ÷ 8 is 78 with a remainder of 5. So the least significant digit is 5.

Take the quotient 78 and divide by 8. 8*9 = 72, so 78 ÷ 8 is 9 with a remainder of 6. Next digit is 6.

Take the quotient 9 and divide by 8. That's 1 with a remainder of 1. Next digit is 1.

Take the quotient 1 and divide by 8. That's 0 with a remainder of 1. Next digit is 1.

So writing the remainders from last to first, the base 8 representation is 1 1 6 5. So 1165 base 8. Let me check that.

Calculating 1*8^3 + 1*8^2 + 6*8^1 + 5*8^0.

1*512 = 512

1*64 = 64

6*8 = 48

5*1 = 5

Adding them up: 512 + 64 = 576; 576 + 48 = 624; 624 + 5 = 629. Correct. So base 8 is 1165.

Now, I need to compare the digits of base 7 (1556) and base 8 (1165). Let me write them down:

Base 7 digits: 1, 5, 5, 6

Base 8 digits: 1, 1, 6, 5

Wait, let me make sure I have the digits in the correct order. For base 7, the number is 1556, so digits from left to right are 1, 5, 5, 6. Similarly, base 8 is 1165, so digits are 1, 1, 6, 5.

Now, I need to check which digits are common between these two sets.

First, let me list the digits in base 7: {1, 5, 5, 6}. But since digits can repeat, but when considering common digits, we need to check if the same digit exists in both representations, regardless of their positions. However, the problem says "how many of the same digits are found", so I think it's asking for the count of digits that are present in both representations, considering duplicates? Or is it asking for the number of digits that are the same in both representations, considering their positions? Wait, the example given is 121_3 and 413_5 have one digit in common. Let me check that example.

In base 3, 121 has digits 1, 2, 1. In base 5, 413 has digits 4, 1, 3. The common digit is 1. So even though there are two 1s in the first number, it's only counted once? Or is it counted as two? Wait, the example says "one digit in common", so it's considering unique digits. So maybe the question is asking for the number of distinct digits that are common between the two representations.

But let me check the exact wording: "how many of the same digits are found in the base 7 and base 8 representations". The example says that 121_3 and 413_5 have one digit in common. So in that case, the digits in base 3 are 1, 2, 1 and in base 5 are 4, 1, 3. The common digit is 1. So even though there are two 1s in base 3, it's only counted once. Therefore, the answer is the number of distinct digits that are common between the two representations.

But let me make sure. Suppose if there was a digit that appears in both representations, but multiple times. For example, if base 7 had two 1s and base 8 had one 1, would that count as one or two? The example suggests that it's counted once. So the answer is the number of distinct digits that are present in both representations.

Therefore, for our problem, we need to find the intersection of the sets of digits in base 7 and base 8 representations.

So let's list the digits in base 7: 1, 5, 5, 6. So the set is {1, 5, 6}.

Digits in base 8: 1, 1, 6, 5. The set is {1, 5, 6} as well.

Wait, so both sets have digits {1, 5, 6}. Therefore, the intersection is all three digits. Therefore, the number of same digits is 3? But wait, let me check again.

Wait, base 7 digits: 1, 5, 5, 6. So digits are 1, 5, 6.

Base 8 digits: 1, 1, 6, 5. So digits are 1, 6, 5.

Therefore, the digits in common are 1, 5, 6. So three digits.

But wait, the example had digits 1, 2, 1 and 4, 1, 3. The common digit is 1, so only one. So in that case, even though there are two 1s in the first number, it's only counted once. Therefore, in our case, since all three digits are common, the answer is 3?

But let me check again. Wait, maybe the problem is asking for the number of digits that are the same in both representations, considering their positions? For example, if a digit is in the same position in both representations, then it's counted? But the example given is different. Let me check the example again.

The example says: 121_3 and 413_5 have one digit in common. Let me write them:

121_3 is digits [1, 2, 1]

413_5 is digits [4, 1, 3]

Comparing each position:

First digit: 1 vs 4 – different

Second digit: 2 vs 1 – different

Third digit: 1 vs 3 – different

So none of the digits are the same in the same position? But the example says they have one digit in common. Wait, that contradicts. Wait, maybe the example is not about same position?

Wait, maybe the example is about digits that are present in both numbers, regardless of position. So in 121_3, digits are 1, 2, 1. In 413_5, digits are 4, 1, 3. The common digit is 1. So even though there are two 1s in the first number, it's only counted once. Therefore, the answer is 1.

Therefore, the problem is asking for the number of distinct digits that are present in both representations.

Therefore, in our case, base 7 digits are {1, 5, 6} and base 8 digits are {1, 5, 6}, so the intersection is {1, 5, 6}, so three digits. Therefore, the answer is 3?

But let me check again. Wait, maybe I made a mistake in converting the numbers. Let me double-check the base 7 and base 8 representations.

Starting with base 7:

629 divided by 7: 7*89 = 623, remainder 6. Then 89 divided by 7: 7*12 = 84, remainder 5. Then 12 divided by 7: 7*1 = 7, remainder 5. Then 1 divided by 7: 0 remainder 1. So digits are 1, 5, 5, 6. So 1556 base 7. Correct.

Base 8: 629 divided by 8: 8*78 = 624, remainder 5. Then 78 divided by 8: 8*9 = 72, remainder 6. Then 9 divided by 8: 1 remainder 1. Then 1 divided by 8: 0 remainder 1. So digits are 1, 1, 6, 5. So 1165 base 8. Correct.

So digits in base 7: 1, 5, 5, 6. So digits are 1, 5, 6.

Digits in base 8: 1, 1, 6, 5. So digits are 1, 6, 5.

Therefore, the digits in common are 1, 5, 6. So three digits.

But wait, the problem says "how many of the same digits are found in the base 7 and base 8 representations". So maybe it's not about distinct digits, but about the count of digits that are same in both representations, considering multiplicity? For example, if a digit appears twice in one and once in the other, does it count as two?

But the example given is 121_3 and 413_5. If we consider multiplicity, then in 121_3, there are two 1s and in 413_5, there is one 1. So total same digits would be two? But the example says one. Therefore, the example suggests that it's about distinct digits.

Therefore, the answer is 3.

But let me check again. Maybe I need to check if there are any digits that are same in both representations, but maybe considering the digits in the numbers. Wait, for example, in base 7, digits can be 0-6, and in base 8, digits can be 0-7. However, the digits in the numbers are only those that are present.

But in our case, both numbers have digits 1, 5, 6. So all three digits are common. Therefore, the answer is 3.

But let me check again. Wait, maybe I made a mistake in the conversion? Let me check base 7 again.

Wait, 1*7^3 is 343, 5*7^2 is 5*49=245, 5*7^1=35, 6*1=6. 343+245=588, 588+35=623, 623+6=629. Correct.

Base 8: 1*8^3=512, 1*8^2=64, 6*8=48, 5*1=5. 512+64=576, 576+48=624, 624+5=629. Correct.

So the digits are correct.

Therefore, digits in base 7: 1, 5, 5, 6.

Digits in base 8: 1, 1, 6, 5.

So the digits present in both are 1, 5, 6. Therefore, three digits.

But wait, the problem says "how many of the same digits are found in the base 7 and base 8 representations". Maybe the question is asking for the number of digits that are the same in both representations, meaning that if a digit is present in both, how many times does it appear? For example, if digit 1 is present in both, how many times? But the example says that 121_3 and 413_5 have one digit in common. In that case, even though there are two 1s in the first number, it's only counted once. Therefore, the answer is the number of distinct digits that are common.

Therefore, the answer is 3.

But let me check again. Maybe the problem is from a source where "same digits" refers to digits that are in the same position? For example, if a digit is in the same position in both representations, then it's counted.

But in that case, let's check.

Base 7: 1 5 5 6

Base 8: 1 1 6 5

Comparing positions:

Position 1: 1 vs 1 – same

Position 2: 5 vs 1 – different

Position 3: 5 vs 6 – different

Position 4: 6 vs 5 – different

So only one digit is same in the same position. But the example given in the problem says that 121_3 and 413_5 have one digit in common. However, in that example, if we check positions:

121_3: positions [1, 2, 1]

413_5: positions [4, 1, 3]

Comparing positions:

Position 1: 1 vs 4 – different

Position 2: 2 vs 1 – different

Position 3: 1 vs 3 – different

So none of the digits are same in the same position. But the example says they have one digit in common. Therefore, the example is not about same position. Therefore, the problem is about digits that are present in both representations, regardless of position.

Therefore, the answer is 3.

But let me check again. Maybe the problem is from a source where digits are considered with multiplicity? For example, if a digit appears multiple times in both representations, how does that count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, even though there are two 1s in the first number, it's only counted once. Therefore, the answer is the number of distinct digits that are common.

Therefore, in our case, since the digits are {1, 5, 6} in both, the answer is 3.

But let me check once more. Maybe I made a mistake in the digits.

Base 7: 1556. So digits are 1, 5, 5, 6.

Base 8: 1165. So digits are 1, 1, 6, 5.

So digits in base 7: 1, 5, 6.

Digits in base 8: 1, 5, 6.

Therefore, the intersection is {1, 5, 6}, so three digits.

Therefore, the answer is 3.

But wait, the problem says "how many of the same digits are found in the base 7 and base 8 representations". Maybe the question is asking for the count of digits that are same in both representations, considering the digits in the numbers. For example, if a digit is present in both representations, how many times does it appear?

But in that case, for example, digit 1 appears in both representations. In base 7, it appears once, in base 8, it appears twice. So how does that count?

But the example given is 121_3 and 413_5. In that case, digit 1 is present in both. In 121_3, it's present twice, in 413_5, once. If we count the number of same digits considering multiplicity, it would be 2 (from base 3) and 1 (from base 5), but the example says one. Therefore, the example suggests that it's not considering multiplicity.

Therefore, the answer is 3.

But maybe I need to check if there is a mistake in the problem. Wait, let me check once again.

Wait, maybe the problem is not about the digits in the numbers, but about the digits in the representations. For example, if the base 7 representation has digits 1, 5, 5, 6 and base 8 has 1, 1, 6, 5. So the digits in base 7 are 1, 5, 6 (with 5 repeated). The digits in base 8 are 1, 6, 5 (with 1 repeated). So the digits that are common are 1, 5, 6. So three digits.

Therefore, the answer is 3.

But maybe the problem is in the way digits are considered. For example, if the base 7 representation is 1556, which is four digits, and base 8 is 1165, which is also four digits. Maybe the problem is asking for the number of digits that are same in both representations, considering their positions? For example, if a digit is same in both representations at the same position, then it's counted.

In that case, let's check:

Base 7: positions 1, 2, 3, 4: digits 1, 5, 5, 6

Base 8: positions 1, 2, 3, 4: digits 1, 1, 6, 5

Comparing position by position:

Position 1: 1 vs 1 – same

Position 2: 5 vs 1 – different

Position 3: 5 vs 6 – different

Position 4: 6 vs 5 – different

So only one digit is same in the same position.

But the example given in the problem says that 121_3 and 413_5 have one digit in common. However, in that case, if we check positions:

121_3: positions 1, 2, 3: 1, 2, 1

413_5: positions 1, 2, 3: 4, 1, 3

Comparing positions:

Position 1: 1 vs 4 – different

Position 2: 2 vs 1 – different

Position 3: 1 vs 3 – different

So none of the digits are same in the same position. But the example says they have one digit in common. Therefore, the example is not about same position. Therefore, the problem is about digits that are present in both representations, regardless of position.

Therefore, the answer is 3.

But maybe the problem is from a source where digits are considered with multiplicity? For example, if a digit is present in both representations, how many times does it appear?

But in that case, for example, digit 1 appears once in base 7 and twice in base 8. So total same digits would be 1 (from base 7) and 2 (from base 8). But how to count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, digit 1 is present in both. In base 3, it's present twice, in base 5 once. If we count the number of same digits considering multiplicity, it would be 1 (the minimum of the counts?), but the example says one.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, regardless of how many times they appear. For example, if a digit is present in both, it's counted once.

Therefore, the answer is 3.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, considering the digits in the numbers. For example, if a digit is present in both representations, how many times does it appear?

But in that case, for example, digit 1 is present once in base 7 and twice in base 8. So how does that count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, digit 1 is present in both. So even though there are two 1s in base 3 and one in base 5, it's counted once. Therefore, the answer is the number of distinct digits that are common.

Therefore, the answer is 3.

But to be thorough, let me check if there is any other interpretation.

Another way: maybe the problem is asking for the number of digits that are same in both representations, meaning that if a digit is present in both representations, it's counted once, regardless of how many times it appears.

Therefore, since digits 1, 5, 6 are present in both, the answer is 3.

Alternatively, if the problem is asking for the number of digits that are same in both representations, considering the digits in the numbers, but not counting duplicates. For example, if a digit is present in both, it's counted once.

Therefore, the answer is 3.

Therefore, I think the answer is 3.

But let me check once again.

Wait, maybe I made a mistake in the base 7 and base 8 representations. Let me check again.

For base 7:

629 divided by 7: 7*89 = 623, remainder 6.

89 divided by 7: 7*12 = 84, remainder 5.

12 divided by 7: 7*1 = 7, remainder 5.

1 divided by 7: 0 remainder 1.

So digits are 1, 5, 5, 6.

For base 8:

629 divided by 8: 8*78 = 624, remainder 5.

78 divided by 8: 8*9 = 72, remainder 6.

9 divided by 8: 1 remainder 1.

1 divided by 8: 0 remainder 1.

Digits are 1, 1, 6, 5.

Yes, that's correct.

So digits in base 7: 1, 5, 5, 6.

Digits in base 8: 1, 1, 6, 5.

So digits present in both are 1, 5, 6.

Therefore, the answer is 3.

But maybe the problem is in the way digits are considered. For example, if leading zeros are considered? But in base 7 and base 8 representations, there are no leading zeros.

Alternatively, maybe the problem is considering digits that are same in both representations, but not necessarily present in both. Wait, no.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, meaning that if a digit is present in both representations, how many times does it appear? For example, digit 1 appears once in base 7 and twice in base 8. So total same digits would be 1 (from base 7) and 2 (from base 8). But how to count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, digit 1 is present in both. If we count the number of same digits considering multiplicity, it would be 2 (from base 3) and 1 (from base 5). But the example says one. Therefore, the example suggests that it's not considering multiplicity.

Therefore, the answer is 3.

Therefore, I think the answer is 3.

But to be 100% sure, let me check if there is any other possible mistake.

Wait, maybe the problem is asking for the number of digits that are same in both representations, meaning that if a digit is present in both representations, it's counted once, even if it appears multiple times.

Yes, that's the example. Therefore, answer is 3.

Therefore, I think the answer is 3.

**Final Answer**
\boxed{3}

To determine how many of the same digits are found in the **base 7** and **base 8** representations of the decimal number **629**, we proceed in two steps: first, convert the number to both bases, and second, compare the digits in those representations.

---

### Step 1: Convert 629 from Base 10 to Base 7

We use the **division-remainder method**:

- $ 629 \div 7 = 89 $ remainder **6**
- $ 89 \div 7 = 12 $ remainder **5**
- $ 12 \div 7 = 1 $ remainder **5**
- $ 1 \div 7 = 0 $ remainder **1**

Reading the remainders from bottom to top, the base 7 representation is:

$$
629_{10} = 1556_7
$$

Digits in base 7: **1, 5, 5, 6**

---

### Step 2: Convert 629 from Base 10 to Base 8

Again, using the division-remainder method:

- $ 629 \div 8 = 78 $ remainder **5**
- $ 78 \div 8 = 9 $ remainder **6**
- $ 9 \div 8 = 1 $ remainder **1**
- $ 1 \div 8 = 0 $ remainder **1**

Reading the remainders from bottom to top, the base 8 representation is:

$$
629_{10} = 1165_8
$$

Digits in base 8: **1, 1, 6, 5**

---

### Step 3: Compare the Digits

Now, we compare the **sets of digits** in both representations:

- **Base 7 digits**: {1, 5, 6}
- **Base 8 digits**: {1, 5, 6}

The **intersection** of these two sets is:

$$
\{1, 5, 6\}
$$

This means that the digits **1**, **5**, and **6** are **common** to both base 7 and base 8 representations.

---

### Final Interpretation

The question asks: *"How many of the same digits are found in the base 7 and base 8 representations?"*

From the example provided in the problem, it's clear that the question is asking for the **number of distinct digits** that are **common** to both representations, **regardless of how many times they appear** in each.

Thus, the answer is the number of **distinct digits** that appear in both representations.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many of the same digits are found in the base 7 and base 8 representations of the decimal number 629. Let me start by recalling how to convert numbers between different bases.

First, I should convert 629 from base 10 to base 7 and then to base 8. Once I have both representations, I can compare the digits to see which ones are common. The question is asking for the count of digits that are the same in both representations.

Starting with base 7 conversion. To convert a decimal number to another base, I can use the division-remainder method. Let me try that.

So, for base 7:

Divide 629 by 7 and record the remainders.

First division: 629 ÷ 7. Let me calculate that. 7*89 is 623, right? 7*90 is 630, which is 1 more than 629. So 629 ÷ 7 is 89 with a remainder of 6. So the least significant digit (rightmost digit) is 6.

Now take the quotient, which is 89, and divide by 7 again. 7*12 is 84, so 89 ÷ 7 is 12 with a remainder of 5. So the next digit is 5.

Take the quotient 12 and divide by 7. 7*1 is 7, so 12 ÷ 7 is 1 with a remainder of 5. Next digit is 5.

Take the quotient 1 and divide by 7. That's 0 with a remainder of 1. So the next digit is 1.

Since the quotient is now 0, we stop. So writing the remainders from last to first, the base 7 representation is 1 5 5 6. So 1556 base 7. Let me check that.

Let me verify: 1*7^3 + 5*7^2 + 5*7^1 + 6*7^0.

Calculating each term:

1*343 = 343

5*49 = 245

5*7 = 35

6*1 = 6

Adding them up: 343 + 245 = 588; 588 + 35 = 623; 623 + 6 = 629. Perfect, that's correct. So base 7 is 1556.

Now moving on to base 8. Let me do the same process for base 8.

Convert 629 to base 8.

Divide 629 by 8. Let me calculate 8*78 = 624. So 629 ÷ 8 is 78 with a remainder of 5. So the least significant digit is 5.

Take the quotient 78 and divide by 8. 8*9 = 72, so 78 ÷ 8 is 9 with a remainder of 6. Next digit is 6.

Take the quotient 9 and divide by 8. That's 1 with a remainder of 1. Next digit is 1.

Take the quotient 1 and divide by 8. That's 0 with a remainder of 1. Next digit is 1.

So writing the remainders from last to first, the base 8 representation is 1 1 6 5. So 1165 base 8. Let me check that.

Calculating 1*8^3 + 1*8^2 + 6*8^1 + 5*8^0.

1*512 = 512

1*64 = 64

6*8 = 48

5*1 = 5

Adding them up: 512 + 64 = 576; 576 + 48 = 624; 624 + 5 = 629. Correct. So base 8 is 1165.

Now, I need to compare the digits of base 7 (1556) and base 8 (1165). Let me write them down:

Base 7 digits: 1, 5, 5, 6

Base 8 digits: 1, 1, 6, 5

Wait, let me make sure I have the digits in the correct order. For base 7, the number is 1556, so digits from left to right are 1, 5, 5, 6. Similarly, base 8 is 1165, so digits are 1, 1, 6, 5.

Now, I need to check which digits are common between these two sets.

First, let me list the digits in base 7: {1, 5, 5, 6}. But since digits can repeat, but when considering common digits, we need to check if the same digit exists in both representations, regardless of their positions. However, the problem says "how many of the same digits are found", so I think it's asking for the count of digits that are present in both representations, considering duplicates? Or is it asking for the number of digits that are the same in both representations, considering their positions? Wait, the example given is 121_3 and 413_5 have one digit in common. Let me check that example.

In base 3, 121 has digits 1, 2, 1. In base 5, 413 has digits 4, 1, 3. The common digit is 1. So even though there are two 1s in the first number, it's only counted once? Or is it counted as two? Wait, the example says "one digit in common", so it's considering unique digits. So maybe the question is asking for the number of distinct digits that are common between the two representations.

But let me check the exact wording: "how many of the same digits are found in the base 7 and base 8 representations". The example says that 121_3 and 413_5 have one digit in common. So in that case, the digits in base 3 are 1, 2, 1 and in base 5 are 4, 1, 3. The common digit is 1. So even though there are two 1s in base 3, it's only counted once. Therefore, the answer is the number of distinct digits that are common between the two representations.

But let me make sure. Suppose if there was a digit that appears in both representations, but multiple times. For example, if base 7 had two 1s and base 8 had one 1, would that count as one or two? The example suggests that it's counted once. So the answer is the number of distinct digits that are present in both representations.

Therefore, for our problem, we need to find the intersection of the sets of digits in base 7 and base 8 representations.

So let's list the digits in base 7: 1, 5, 5, 6. So the set is {1, 5, 6}.

Digits in base 8: 1, 1, 6, 5. The set is {1, 5, 6} as well.

Wait, so both sets have digits {1, 5, 6}. Therefore, the intersection is all three digits. Therefore, the number of same digits is 3? But wait, let me check again.

Wait, base 7 digits: 1, 5, 5, 6. So digits are 1, 5, 6.

Base 8 digits: 1, 1, 6, 5. So digits are 1, 6, 5.

Therefore, the digits in common are 1, 5, 6. So three digits.

But wait, the example had digits 1, 2, 1 and 4, 1, 3. The common digit is 1, so only one. So in that case, even though there are two 1s in the first number, it's only counted once. Therefore, in our case, since all three digits are common, the answer is 3?

But let me check again. Wait, maybe the problem is asking for the number of digits that are the same in both representations, considering their positions? For example, if a digit is in the same position in both representations, then it's counted? But the example given is different. Let me check the example again.

The example says: 121_3 and 413_5 have one digit in common. Let me write them:

121_3 is digits [1, 2, 1]

413_5 is digits [4, 1, 3]

Comparing each position:

First digit: 1 vs 4 – different

Second digit: 2 vs 1 – different

Third digit: 1 vs 3 – different

So none of the digits are the same in the same position? But the example says they have one digit in common. Wait, that contradicts. Wait, maybe the example is not about same position?

Wait, maybe the example is about digits that are present in both numbers, regardless of position. So in 121_3, digits are 1, 2, 1. In 413_5, digits are 4, 1, 3. The common digit is 1. So even though there are two 1s in the first number, it's only counted once. Therefore, the answer is 1.

Therefore, the problem is asking for the number of distinct digits that are present in both representations.

Therefore, in our case, base 7 digits are {1, 5, 6} and base 8 digits are {1, 5, 6}, so the intersection is {1, 5, 6}, so three digits. Therefore, the answer is 3?

But let me check again. Wait, maybe I made a mistake in converting the numbers. Let me double-check the base 7 and base 8 representations.

Starting with base 7:

629 divided by 7: 7*89 = 623, remainder 6. Then 89 divided by 7: 7*12 = 84, remainder 5. Then 12 divided by 7: 7*1 = 7, remainder 5. Then 1 divided by 7: 0 remainder 1. So digits are 1, 5, 5, 6. So 1556 base 7. Correct.

Base 8: 629 divided by 8: 8*78 = 624, remainder 5. Then 78 divided by 8: 8*9 = 72, remainder 6. Then 9 divided by 8: 1 remainder 1. Then 1 divided by 8: 0 remainder 1. So digits are 1, 1, 6, 5. So 1165 base 8. Correct.

So digits in base 7: 1, 5, 5, 6. So digits are 1, 5, 6.

Digits in base 8: 1, 1, 6, 5. So digits are 1, 6, 5.

Therefore, the digits in common are 1, 5, 6. So three digits.

But wait, the problem says "how many of the same digits are found in the base 7 and base 8 representations". So maybe it's not about distinct digits, but about the count of digits that are same in both representations, considering multiplicity? For example, if a digit appears twice in one and once in the other, does it count as two?

But the example given is 121_3 and 413_5. If we consider multiplicity, then in 121_3, there are two 1s and in 413_5, there is one 1. So total same digits would be two? But the example says one. Therefore, the example suggests that it's about distinct digits.

Therefore, the answer is 3.

But let me check again. Maybe I need to check if there are any digits that are same in both representations, but maybe considering the digits in the numbers. Wait, for example, in base 7, digits can be 0-6, and in base 8, digits can be 0-7. However, the digits in the numbers are only those that are present.

But in our case, both numbers have digits 1, 5, 6. So all three digits are common. Therefore, the answer is 3.

But let me check again. Wait, maybe I made a mistake in the conversion? Let me check base 7 again.

Wait, 1*7^3 is 343, 5*7^2 is 5*49=245, 5*7^1=35, 6*1=6. 343+245=588, 588+35=623, 623+6=629. Correct.

Base 8: 1*8^3=512, 1*8^2=64, 6*8=48, 5*1=5. 512+64=576, 576+48=624, 624+5=629. Correct.

So the digits are correct.

Therefore, digits in base 7: 1, 5, 5, 6.

Digits in base 8: 1, 1, 6, 5.

So the digits present in both are 1, 5, 6. Therefore, three digits.

But wait, the problem says "how many of the same digits are found in the base 7 and base 8 representations". Maybe the question is asking for the number of digits that are the same in both representations, meaning that if a digit is present in both, how many times does it appear? For example, if digit 1 is present in both, how many times? But the example says that 121_3 and 413_5 have one digit in common. In that case, even though there are two 1s in the first number, it's only counted once. Therefore, the answer is the number of distinct digits that are common.

Therefore, the answer is 3.

But let me check again. Maybe the problem is from a source where "same digits" refers to digits that are in the same position? For example, if a digit is in the same position in both representations, then it's counted.

But in that case, let's check.

Base 7: 1 5 5 6

Base 8: 1 1 6 5

Comparing positions:

Position 1: 1 vs 1 – same

Position 2: 5 vs 1 – different

Position 3: 5 vs 6 – different

Position 4: 6 vs 5 – different

So only one digit is same in the same position. But the example given in the problem says that 121_3 and 413_5 have one digit in common. However, in that example, if we check positions:

121_3: positions [1, 2, 1]

413_5: positions [4, 1, 3]

Comparing positions:

Position 1: 1 vs 4 – different

Position 2: 2 vs 1 – different

Position 3: 1 vs 3 – different

So none of the digits are same in the same position. But the example says they have one digit in common. Therefore, the example is not about same position. Therefore, the problem is about digits that are present in both representations, regardless of position.

Therefore, the answer is 3.

But let me check again. Maybe the problem is from a source where digits are considered with multiplicity? For example, if a digit appears multiple times in both representations, how does that count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, even though there are two 1s in the first number, it's only counted once. Therefore, the answer is the number of distinct digits that are common.

Therefore, in our case, since the digits are {1, 5, 6} in both, the answer is 3.

But let me check once more. Maybe I made a mistake in the digits.

Base 7: 1556. So digits are 1, 5, 5, 6.

Base 8: 1165. So digits are 1, 1, 6, 5.

So digits in base 7: 1, 5, 6.

Digits in base 8: 1, 5, 6.

Therefore, the intersection is {1, 5, 6}, so three digits.

Therefore, the answer is 3.

But wait, the problem says "how many of the same digits are found in the base 7 and base 8 representations". Maybe the question is asking for the count of digits that are same in both representations, considering the digits in the numbers. For example, if a digit is present in both representations, how many times does it appear?

But in that case, for example, digit 1 appears in both representations. In base 7, it appears once, in base 8, it appears twice. So how does that count?

But the example given is 121_3 and 413_5. In that case, digit 1 is present in both. In 121_3, it's present twice, in 413_5, once. If we count the number of same digits considering multiplicity, it would be 2 (from base 3) and 1 (from base 5), but the example says one. Therefore, the example suggests that it's not considering multiplicity.

Therefore, the answer is 3.

But maybe I need to check if there is a mistake in the problem. Wait, let me check once again.

Wait, maybe the problem is not about the digits in the numbers, but about the digits in the representations. For example, if the base 7 representation has digits 1, 5, 5, 6 and base 8 has 1, 1, 6, 5. So the digits in base 7 are 1, 5, 6 (with 5 repeated). The digits in base 8 are 1, 6, 5 (with 1 repeated). So the digits that are common are 1, 5, 6. So three digits.

Therefore, the answer is 3.

But maybe the problem is in the way digits are considered. For example, if the base 7 representation is 1556, which is four digits, and base 8 is 1165, which is also four digits. Maybe the problem is asking for the number of digits that are same in both representations, considering their positions? For example, if a digit is same in both representations at the same position, then it's counted.

In that case, let's check:

Base 7: positions 1, 2, 3, 4: digits 1, 5, 5, 6

Base 8: positions 1, 2, 3, 4: digits 1, 1, 6, 5

Comparing position by position:

Position 1: 1 vs 1 – same

Position 2: 5 vs 1 – different

Position 3: 5 vs 6 – different

Position 4: 6 vs 5 – different

So only one digit is same in the same position.

But the example given in the problem says that 121_3 and 413_5 have one digit in common. However, in that case, if we check positions:

121_3: positions 1, 2, 3: 1, 2, 1

413_5: positions 1, 2, 3: 4, 1, 3

Comparing positions:

Position 1: 1 vs 4 – different

Position 2: 2 vs 1 – different

Position 3: 1 vs 3 – different

So none of the digits are same in the same position. But the example says they have one digit in common. Therefore, the example is not about same position. Therefore, the problem is about digits that are present in both representations, regardless of position.

Therefore, the answer is 3.

But maybe the problem is from a source where digits are considered with multiplicity? For example, if a digit is present in both representations, how many times does it appear?

But in that case, for example, digit 1 appears once in base 7 and twice in base 8. So total same digits would be 1 (from base 7) and 2 (from base 8). But how to count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, digit 1 is present in both. In base 3, it's present twice, in base 5 once. If we count the number of same digits considering multiplicity, it would be 1 (the minimum of the counts?), but the example says one.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, regardless of how many times they appear. For example, if a digit is present in both, it's counted once.

Therefore, the answer is 3.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, considering the digits in the numbers. For example, if a digit is present in both representations, how many times does it appear?

But in that case, for example, digit 1 is present once in base 7 and twice in base 8. So how does that count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, digit 1 is present in both. So even though there are two 1s in base 3 and one in base 5, it's counted once. Therefore, the answer is the number of distinct digits that are common.

Therefore, the answer is 3.

But to be thorough, let me check if there is any other interpretation.

Another way: maybe the problem is asking for the number of digits that are same in both representations, meaning that if a digit is present in both representations, it's counted once, regardless of how many times it appears.

Therefore, since digits 1, 5, 6 are present in both, the answer is 3.

Alternatively, if the problem is asking for the number of digits that are same in both representations, considering the digits in the numbers, but not counting duplicates. For example, if a digit is present in both, it's counted once.

Therefore, the answer is 3.

Therefore, I think the answer is 3.

But let me check once again.

Wait, maybe I made a mistake in the base 7 and base 8 representations. Let me check again.

For base 7:

629 divided by 7: 7*89 = 623, remainder 6.

89 divided by 7: 7*12 = 84, remainder 5.

12 divided by 7: 7*1 = 7, remainder 5.

1 divided by 7: 0 remainder 1.

So digits are 1, 5, 5, 6.

For base 8:

629 divided by 8: 8*78 = 624, remainder 5.

78 divided by 8: 8*9 = 72, remainder 6.

9 divided by 8: 1 remainder 1.

1 divided by 8: 0 remainder 1.

Digits are 1, 1, 6, 5.

Yes, that's correct.

So digits in base 7: 1, 5, 5, 6.

Digits in base 8: 1, 1, 6, 5.

So digits present in both are 1, 5, 6.

Therefore, the answer is 3.

But maybe the problem is in the way digits are considered. For example, if leading zeros are considered? But in base 7 and base 8 representations, there are no leading zeros.

Alternatively, maybe the problem is considering digits that are same in both representations, but not necessarily present in both. Wait, no.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, meaning that if a digit is present in both representations, how many times does it appear? For example, digit 1 appears once in base 7 and twice in base 8. So total same digits would be 1 (from base 7) and 2 (from base 8). But how to count?

But the example says that 121_3 and 413_5 have one digit in common. In that case, digit 1 is present in both. If we count the number of same digits considering multiplicity, it would be 2 (from base 3) and 1 (from base 5). But the example says one. Therefore, the example suggests that it's not considering multiplicity.

Therefore, the answer is 3.

Therefore, I think the answer is 3.

But to be 100% sure, let me check if there is any other possible mistake.

Wait, maybe the problem is asking for the number of digits that are same in both representations, meaning that if a digit is present in both representations, it's counted once, even if it appears multiple times.

Yes, that's the example. Therefore, answer is 3.

Therefore, I think the answer is 3.

**Final Answer**
\boxed{3}

To determine how many of the same digits are found in the **base 7** and **base 8** representations of the decimal number **629**, we proceed in two steps: first, convert the number to both bases, and second, compare the digits in those representations.

---

### Step 1: Convert 629 from Base 10 to Base 7

We use the **division-remainder method**:

- $ 629 \div 7 = 89 $ remainder **6**
- $ 89 \div 7 = 12 $ remainder **5**
- $ 12 \div 7 = 1 $ remainder **5**
- $ 1 \div 7 = 0 $ remainder **1**

Reading the remainders from bottom to top, the base 7 representation is:

$$
629_{10} = 1556_7
$$

Digits in base 7: **1, 5, 5, 6**

---

### Step 2: Convert 629 from Base 10 to Base 8

Again, using the division-remainder method:

- $ 629 \div 8 = 78 $ remainder **5**
- $ 78 \div 8 = 9 $ remainder **6**
- $ 9 \div 8 = 1 $ remainder **1**
- $ 1 \div 8 = 0 $ remainder **1**

Reading the remainders from bottom to top, the base 8 representation is:

$$
629_{10} = 1165_8
$$

Digits in base 8: **1, 1, 6, 5**

---

### Step 3: Compare the Digits

Now, we compare the **sets of digits** in both representations:

- **Base 7 digits**: {1, 5, 6}
- **Base 8 digits**: {1, 5, 6}

The **intersection** of these two sets is:

$$
\{1, 5, 6\}
$$

This means that the digits **1**, **5**, and **6** are **common** to both base 7 and base 8 representations.

---

### Final Interpretation

The question asks: *"How many of the same digits are found in the base 7 and base 8 representations?"*

From the example provided in the problem, it's clear that the question is asking for the **number of distinct digits** that are **common** to both representations, **regardless of how many times they appear** in each.

Thus, the answer is the number of **distinct digits** that appear in both representations.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
First, let us convert $629_{10}$ to each of the two bases. To convert to base 7, we realize $7^{4}>629_{10}>7^{3}$. So, we can tell that $629_{10}$ in base seven will have four digits. $7^{3}=343$, which can go into 629 only one time at most, leaving $629-1\cdot343 = 286$ for the next three digits. $7^{2}=49$ goes into 286 five times at most, leaving us with $286-5\cdot49 = 41$. Then, $7^{1}=7$ goes into 41 five times at most, leaving $41-5\cdot7 = 6$ for the ones digit. All together, the base seven equivalent of $629_{10}$ is $1556_{7}$.

To convert to base 8, we realize similarly that $8^{4}>629_{10}>8^{3}$. So, we can tell that $629_{10}$ in base eight will have four digits. $8^{3}=512$, which can go into 629 only one time at most, leaving $629-1\cdot512 = 117$ for the next three digits. $8^{2}=64$ goes into 117 one time at most, leaving us with $117-1\cdot64 = 53$. Then, $8^{1}=8$ goes into 53 six times at most, leaving $53-6\cdot8 = 5$ for the ones digit. All together, the base eight equivalent of $629_{10}$ is $1165_{8}$.

Finally, comparing $1556_{7}$ and $1165_{8}$, we find that digits 1, 5, and 6 are present in both numbers.  Thus, there are $\boxed{3}$ digits in common.


Explanation:
Okay, so I need to figure out how many ways 3 people can be assigned to 5 different offices if no one shares an office. Each person gets exactly one office. Let me think about this step by step.

First, let me make sure I understand the problem correctly. There are 5 offices, and 3 people. Each person must have their own office, so no two people can be in the same office. So, essentially, I need to assign each of the 3 people to a unique office among the 5 available.

Hmm, this seems like a permutation problem because the order in which we assign the offices might matter. Wait, actually, maybe it's a combination problem? Let me think. If the offices are different, then assigning person A to office 1 and person B to office 2 is different from assigning person A to office 2 and person B to office 1. So, the order does matter here because each office is distinct. Therefore, permutations might be the right approach.

Alternatively, I can think of this as a selection problem where I first choose 3 offices out of 5 and then assign the people to those offices. Let me explore both approaches to see if they lead to the same answer.

First approach: Assigning each person to an office one by one.

For the first person, there are 5 offices to choose from. Once the first person has been assigned an office, there are 4 offices left for the second person. Then, for the third person, there are 3 offices remaining. So, the total number of ways would be 5 * 4 * 3. Let me calculate that: 5 * 4 is 20, and 20 * 3 is 60. So, 60 ways.

Second approach: Choosing offices first and then assigning people.

First, I need to choose 3 distinct offices out of 5. The number of ways to choose 3 offices from 5 is given by the combination formula C(5,3). Then, for each such selection of offices, I need to assign the 3 people to these offices. Since the offices are different, the assignment is a permutation of the people. So, the number of permutations of 3 people is 3! (which is 6). Therefore, the total number of ways would be C(5,3) * 3!.

Let me compute that. C(5,3) is 10, and 10 * 6 is 60. So, same answer as before. That's reassuring.

Alternatively, another way to think about this is using permutations directly. The number of permutations of 5 offices taken 3 at a time is P(5,3) which is calculated as 5! / (5-3)! = 5! / 2! = (120) / 2 = 60. Yep, same result again. So, all three methods give me 60. That seems consistent.

Wait, let me just make sure I didn't make a mistake in interpreting the problem. The problem says "no one shares an office," so each person must have their own office. So, we are assigning each person to a unique office. Therefore, the answer is indeed 5 * 4 * 3 = 60.

Alternatively, if the offices were identical, then it would be a combination problem, but since the offices are different, the assignments are different. For example, if office A is different from office B, then assigning person 1 to office A and person 2 to office B is different from assigning person 1 to office B and person 2 to office A. Therefore, permutations are the right way to go.

Let me think if there's another way someone might approach this problem incorrectly. For instance, someone might think that since there are 5 offices and 3 people, each person has 5 choices, so 5^3 = 125. But that would be the case if multiple people could share offices. However, the problem states that no one shares an office, so that approach is wrong. So, 5^3 is incorrect here because it allows for multiple people in the same office. Therefore, we need to adjust for that by considering permutations instead of combinations with repetition.

Another possible mistake could be thinking that it's a combination problem where order doesn't matter. For example, if we thought that choosing offices 1, 2, 3 is the same as choosing 2, 1, 3, but since the people are distinct, the assignment matters. Therefore, we need to multiply by the number of ways to assign the people to the selected offices, which is 3!.

So, in summary, the answer is 5 * 4 * 3 = 60. Alternatively, using permutations formula P(n, k) = n! / (n - k)! where n = 5 and k = 3, which gives 5! / 2! = 120 / 2 = 60.

Let me check with a smaller example to see if my reasoning holds. Suppose there are 2 offices and 1 person. Then, the number of ways should be 2. Using the formula, P(2,1) = 2! / 1! = 2, which is correct. If there are 2 offices and 2 people, then it's P(2,2) = 2! / 0! = 2 / 1 = 2. Which makes sense: person A in office 1 and person B in office 2, or person B in office 1 and person A in office 2. Correct.

Another example: 3 offices and 2 people. Then, the number of ways should be 3 * 2 = 6. Alternatively, C(3,2) * 2! = 3 * 2 = 6. Correct. So, that seems to hold. Therefore, applying the same logic to 5 offices and 3 people, 5 * 4 * 3 = 60 is correct.

Therefore, I think the answer is 60.

**Final Answer**
The number of ways is \boxed{60}.

To determine the number of ways to assign 3 people to 5 different offices such that **no one shares an office**, we need to consider the constraints and the nature of the problem.

---

### Understanding the Problem

- There are **5 distinct offices**.
- There are **3 distinct people**.
- Each person must be assigned **exactly one office**.
- **No two people can share the same office**.

This means we are looking for the number of **ways to assign 3 distinct people to 3 distinct offices out of 5**, with **no repetition** of offices.

---

### Step-by-Step Reasoning

This is a **permutation problem** because:

- The **order of assignment matters** (i.e., assigning person A to office 1 and person B to office 2 is different from assigning person B to office 1 and person A to office 2).
- The **offices are distinct**, so assigning to different offices results in different configurations.

We can approach this in a few equivalent ways:

---

#### **Approach 1: Sequential Assignment**

- The **first person** has **5 choices** of offices.
- The **second person** has **4 remaining choices**.
- The **third person** has **3 remaining choices**.

So the total number of assignments is:

$$
5 \times 4 \times 3 = 60
$$

---

#### **Approach 2: Combinations and Permutations**

1. **Choose 3 offices out of 5**:
   $$
   \binom{5}{3} = 10
   $$

2. **Assign the 3 people to these 3 offices**:
   Since the people are distinct and the offices are distinct, this is a **permutation** of 3 people:
   $$
   3! = 6
   $$

3. **Total number of assignments**:
   $$
   \binom{5}{3} \times 3! = 10 \times 6 = 60
   $$

---

#### **Approach 3: Permutations of 5 offices taken 3 at a time**

This is a standard permutation formula:

$$
P(5, 3) = \frac{5!}{(5 - 3)!} = \frac{120}{2} = 60
$$

---

### Final Answer

All approaches lead to the same result. Therefore, the number of ways to assign 3 people to 5 different offices, with no one sharing an office, is:

$$
\boxed{60}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many ways 3 people can be assigned to 5 different offices if no one shares an office. Each person gets exactly one office. Let me think about this step by step.

First, let me make sure I understand the problem correctly. There are 5 offices, and 3 people. Each person must have their own office, so no two people can be in the same office. So, essentially, I need to assign each of the 3 people to a unique office among the 5 available.

Hmm, this seems like a permutation problem because the order in which we assign the offices might matter. Wait, actually, maybe it's a combination problem? Let me think. If the offices are different, then assigning person A to office 1 and person B to office 2 is different from assigning person A to office 2 and person B to office 1. So, the order does matter here because each office is distinct. Therefore, permutations might be the right approach.

Alternatively, I can think of this as a selection problem where I first choose 3 offices out of 5 and then assign the people to those offices. Let me explore both approaches to see if they lead to the same answer.

First approach: Assigning each person to an office one by one.

For the first person, there are 5 offices to choose from. Once the first person has been assigned an office, there are 4 offices left for the second person. Then, for the third person, there are 3 offices remaining. So, the total number of ways would be 5 * 4 * 3. Let me calculate that: 5 * 4 is 20, and 20 * 3 is 60. So, 60 ways.

Second approach: Choosing offices first and then assigning people.

First, I need to choose 3 distinct offices out of 5. The number of ways to choose 3 offices from 5 is given by the combination formula C(5,3). Then, for each such selection of offices, I need to assign the 3 people to these offices. Since the offices are different, the assignment is a permutation of the people. So, the number of permutations of 3 people is 3! (which is 6). Therefore, the total number of ways would be C(5,3) * 3!.

Let me compute that. C(5,3) is 10, and 10 * 6 is 60. So, same answer as before. That's reassuring.

Alternatively, another way to think about this is using permutations directly. The number of permutations of 5 offices taken 3 at a time is P(5,3) which is calculated as 5! / (5-3)! = 5! / 2! = (120) / 2 = 60. Yep, same result again. So, all three methods give me 60. That seems consistent.

Wait, let me just make sure I didn't make a mistake in interpreting the problem. The problem says "no one shares an office," so each person must have their own office. So, we are assigning each person to a unique office. Therefore, the answer is indeed 5 * 4 * 3 = 60.

Alternatively, if the offices were identical, then it would be a combination problem, but since the offices are different, the assignments are different. For example, if office A is different from office B, then assigning person 1 to office A and person 2 to office B is different from assigning person 1 to office B and person 2 to office A. Therefore, permutations are the right way to go.

Let me think if there's another way someone might approach this problem incorrectly. For instance, someone might think that since there are 5 offices and 3 people, each person has 5 choices, so 5^3 = 125. But that would be the case if multiple people could share offices. However, the problem states that no one shares an office, so that approach is wrong. So, 5^3 is incorrect here because it allows for multiple people in the same office. Therefore, we need to adjust for that by considering permutations instead of combinations with repetition.

Another possible mistake could be thinking that it's a combination problem where order doesn't matter. For example, if we thought that choosing offices 1, 2, 3 is the same as choosing 2, 1, 3, but since the people are distinct, the assignment matters. Therefore, we need to multiply by the number of ways to assign the people to the selected offices, which is 3!.

So, in summary, the answer is 5 * 4 * 3 = 60. Alternatively, using permutations formula P(n, k) = n! / (n - k)! where n = 5 and k = 3, which gives 5! / 2! = 120 / 2 = 60.

Let me check with a smaller example to see if my reasoning holds. Suppose there are 2 offices and 1 person. Then, the number of ways should be 2. Using the formula, P(2,1) = 2! / 1! = 2, which is correct. If there are 2 offices and 2 people, then it's P(2,2) = 2! / 0! = 2 / 1 = 2. Which makes sense: person A in office 1 and person B in office 2, or person B in office 1 and person A in office 2. Correct.

Another example: 3 offices and 2 people. Then, the number of ways should be 3 * 2 = 6. Alternatively, C(3,2) * 2! = 3 * 2 = 6. Correct. So, that seems to hold. Therefore, applying the same logic to 5 offices and 3 people, 5 * 4 * 3 = 60 is correct.

Therefore, I think the answer is 60.

**Final Answer**
The number of ways is \boxed{60}.

To determine the number of ways to assign 3 people to 5 different offices such that **no one shares an office**, we need to consider the constraints and the nature of the problem.

---

### Understanding the Problem

- There are **5 distinct offices**.
- There are **3 distinct people**.
- Each person must be assigned **exactly one office**.
- **No two people can share the same office**.

This means we are looking for the number of **ways to assign 3 distinct people to 3 distinct offices out of 5**, with **no repetition** of offices.

---

### Step-by-Step Reasoning

This is a **permutation problem** because:

- The **order of assignment matters** (i.e., assigning person A to office 1 and person B to office 2 is different from assigning person B to office 1 and person A to office 2).
- The **offices are distinct**, so assigning to different offices results in different configurations.

We can approach this in a few equivalent ways:

---

#### **Approach 1: Sequential Assignment**

- The **first person** has **5 choices** of offices.
- The **second person** has **4 remaining choices**.
- The **third person** has **3 remaining choices**.

So the total number of assignments is:

$$
5 \times 4 \times 3 = 60
$$

---

#### **Approach 2: Combinations and Permutations**

1. **Choose 3 offices out of 5**:
   $$
   \binom{5}{3} = 10
   $$

2. **Assign the 3 people to these 3 offices**:
   Since the people are distinct and the offices are distinct, this is a **permutation** of 3 people:
   $$
   3! = 6
   $$

3. **Total number of assignments**:
   $$
   \binom{5}{3} \times 3! = 10 \times 6 = 60
   $$

---

#### **Approach 3: Permutations of 5 offices taken 3 at a time**

This is a standard permutation formula:

$$
P(5, 3) = \frac{5!}{(5 - 3)!} = \frac{120}{2} = 60
$$

---

### Final Answer

All approaches lead to the same result. Therefore, the number of ways to assign 3 people to 5 different offices, with no one sharing an office, is:

$$
\boxed{60}
$$

Solution: If we number the three people 1, 2, and 3, there are $5$ offices that person 1 can be assigned to, $4$ offices that person 2 can be assigned to, and $3$ offices that person 3 can be assigned to. This gives us $5 \times 4 \times 3 = \boxed{60}$ ways to assign the three people to offices.


Explanation:
Okay, so I need to find the value of the inverse function of f at 1/5, which is written as f⁻¹(1/5). The function given is f(x) = 2/(x + 1). Hmm, let me recall how to find inverse functions.

First, I think the inverse function essentially undoes what the original function does. So if f(x) takes an x and gives me 2/(x + 1), then the inverse function f⁻¹(y) should take a y and give me back the original x that was used to get y. So, in this case, I need to find the x such that f(x) = 1/5. That makes sense.

Let me write that down. If f(x) = 2/(x + 1), then setting that equal to 1/5:

2/(x + 1) = 1/5

Now, I need to solve this equation for x. Let me do that step by step.

First, cross-multiplying to eliminate the fractions. So, multiplying both sides by (x + 1) and by 5 to get rid of denominators. Let me see:

Multiplying both sides by 5(x + 1):

5 * 2 = 1 * (x + 1)

So that simplifies to:

10 = x + 1

Then, subtracting 1 from both sides:

10 - 1 = x

Which gives:

x = 9

Wait, so does that mean that f⁻¹(1/5) is 9? Let me check my steps again to make sure I didn't make a mistake.

Starting with f(x) = 2/(x + 1). Let me verify if f(9) is indeed 1/5. Plugging x = 9 into f(x):

f(9) = 2/(9 + 1) = 2/10 = 1/5. Yes, that's correct. So that seems to check out.

But just to be thorough, maybe I should go through the process of finding the inverse function properly, step by step, to confirm.

To find the inverse function, I need to switch x and y and then solve for y. Let me try that.

Starting with the original function:

y = 2/(x + 1)

To find the inverse, swap x and y:

x = 2/(y + 1)

Now, solve for y.

Multiply both sides by (y + 1):

x(y + 1) = 2

Expanding the left side:

xy + x = 2

Now, subtract x from both sides:

xy = 2 - x

Then, divide both sides by x:

y = (2 - x)/x

Wait, that seems different from my previous answer. Wait, but hold on, if I have y = (2 - x)/x, then that would be the inverse function? Let me check.

Alternatively, maybe I made a mistake in algebra here. Let me redo that step.

Starting again from x = 2/(y + 1).

Multiply both sides by (y + 1):

x(y + 1) = 2

Which is:

xy + x = 2

Then, subtract x:

xy = 2 - x

Then, divide both sides by x:

y = (2 - x)/x

Alternatively, that can be written as y = 2/x - 1. Let me check that:

(2 - x)/x = 2/x - x/x = 2/x - 1. Yes, that's correct.

So the inverse function f⁻¹(x) is 2/x - 1.

Therefore, to find f⁻¹(1/5), substitute x = 1/5 into this inverse function:

f⁻¹(1/5) = 2/(1/5) - 1

Calculating 2 divided by 1/5 is 2 * 5 = 10. Then subtract 1: 10 - 1 = 9.

So that gives me 9 again. So both methods agree. Therefore, the answer is 9.

But let me think again if there's another way to approach this. Maybe by using function composition?

If f⁻¹(1/5) is the value of x such that f(x) = 1/5, which is exactly what I did first. So solving f(x) = 1/5 gives x = 9, so f⁻¹(1/5) = 9.

Alternatively, maybe I can check if f(f⁻¹(1/5)) = 1/5. Let me see. If f⁻¹(1/5) is 9, then f(9) should be 1/5, which we already checked.

Alternatively, if I take the inverse function f⁻¹(x) = 2/x - 1, then f⁻¹(1/5) = 2/(1/5) - 1 = 10 - 1 = 9.

So all methods lead to the same answer.

Wait, but just to make sure that I didn't make a mistake in finding the inverse function. Let me check with another value. Let's say, for example, if x = 1, then f(1) = 2/(1 + 1) = 1. So f(1) = 1. Then, according to the inverse function, f⁻¹(1) should be 1. Let me check with the inverse function: f⁻¹(1) = 2/1 - 1 = 2 - 1 = 1. Correct.

Another test: Let me take x = 0. Then f(0) = 2/(0 + 1) = 2. So f(0) = 2. Then f⁻¹(2) should be 0. Let me compute f⁻¹(2) using the inverse function: 2/2 - 1 = 1 - 1 = 0. Correct.

Another test: Let me take x = 2. Then f(2) = 2/(2 + 1) = 2/3. Therefore, f⁻¹(2/3) should be 2. Let me compute f⁻¹(2/3): 2/(2/3) - 1 = 3 - 1 = 2. Correct.

So the inverse function seems to be working. Therefore, my calculation for f⁻¹(1/5) = 9 is correct.

Alternatively, maybe I can think about the graph of the function. The original function f(x) = 2/(x + 1) is a hyperbola with a vertical asymptote at x = -1 and a horizontal asymptote at y = 0. The inverse function would be a reflection over the line y = x, so it should also be a hyperbola.

But maybe that's overcomplicating. Since all my algebra checks out and the test cases work, I think I can be confident that the answer is 9.

Wait, but just to make sure, let me try solving the equation again.

Original equation: 2/(x + 1) = 1/5.

Cross-multiplying: 2 * 5 = 1 * (x + 1) => 10 = x + 1 => x = 9.

Yes, that's straightforward.

Alternatively, if I wanted to solve for x in the equation 2/(x + 1) = 1/5, I could also take reciprocals of both sides.

Taking reciprocals: (x + 1)/2 = 5/1 => x + 1 = 10 => x = 9.

Same result.

So, regardless of the method, I get x = 9.

Therefore, I think the answer is definitely 9.

I don't see any mistakes in my reasoning. All steps check out, and multiple methods lead to the same conclusion.

**Final Answer**
The value of $ f^{-1}\left(\frac{1}{5}\right) $ is \boxed{9}.

To find the value of $ f^{-1}\left(\frac{1}{5}\right) $, we begin by understanding what this notation means. The inverse function $ f^{-1}(y) $ gives the value of $ x $ such that $ f(x) = y $. In this case, we are looking for the $ x $ such that:

$$
f(x) = \frac{1}{5}
$$

Given the function:

$$
f(x) = \frac{2}{x + 1}
$$

We set this equal to $ \frac{1}{5} $ and solve for $ x $:

$$
\frac{2}{x + 1} = \frac{1}{5}
$$

---

### Step 1: Cross-Multiply

To eliminate the fractions, we cross-multiply:

$$
2 \cdot 5 = 1 \cdot (x + 1)
$$

$$
10 = x + 1
$$

---

### Step 2: Solve for $ x $

Subtract 1 from both sides:

$$
x = 10 - 1 = 9
$$

---

### Step 3: Verify the Result

To ensure the solution is correct, we substitute $ x = 9 $ back into the original function:

$$
f(9) = \frac{2}{9 + 1} = \frac{2}{10} = \frac{1}{5}
$$

This confirms that $ f(9) = \frac{1}{5} $, so $ f^{-1}\left(\frac{1}{5}\right) = 9 $.

---

### Alternative Method: Finding the Inverse Function

We can also find the inverse function explicitly by switching $ x $ and $ y $ in the original function:

$$
y = \frac{2}{x + 1}
$$

Swap $ x $ and $ y $:

$$
x = \frac{2}{y + 1}
$$

Now solve for $ y $:

$$
x(y + 1) = 2 \Rightarrow xy + x = 2 \Rightarrow xy = 2 - x \Rightarrow y = \frac{2 - x}{x}
$$

So the inverse function is:

$$
f^{-1}(x) = \frac{2 - x}{x}
$$

Now evaluate $ f^{-1}\left(\frac{1}{5}\right) $:

$$
f^{-1}\left(\frac{1}{5}\right) = \frac{2 - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{10 - 1}{5}}{\frac{1}{5}} = \frac{9}{5} \div \frac{1}{5} = 9
$$

---

### Final Answer

$$
\boxed{9}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of the inverse function of f at 1/5, which is written as f⁻¹(1/5). The function given is f(x) = 2/(x + 1). Hmm, let me recall how to find inverse functions.

First, I think the inverse function essentially undoes what the original function does. So if f(x) takes an x and gives me 2/(x + 1), then the inverse function f⁻¹(y) should take a y and give me back the original x that was used to get y. So, in this case, I need to find the x such that f(x) = 1/5. That makes sense.

Let me write that down. If f(x) = 2/(x + 1), then setting that equal to 1/5:

2/(x + 1) = 1/5

Now, I need to solve this equation for x. Let me do that step by step.

First, cross-multiplying to eliminate the fractions. So, multiplying both sides by (x + 1) and by 5 to get rid of denominators. Let me see:

Multiplying both sides by 5(x + 1):

5 * 2 = 1 * (x + 1)

So that simplifies to:

10 = x + 1

Then, subtracting 1 from both sides:

10 - 1 = x

Which gives:

x = 9

Wait, so does that mean that f⁻¹(1/5) is 9? Let me check my steps again to make sure I didn't make a mistake.

Starting with f(x) = 2/(x + 1). Let me verify if f(9) is indeed 1/5. Plugging x = 9 into f(x):

f(9) = 2/(9 + 1) = 2/10 = 1/5. Yes, that's correct. So that seems to check out.

But just to be thorough, maybe I should go through the process of finding the inverse function properly, step by step, to confirm.

To find the inverse function, I need to switch x and y and then solve for y. Let me try that.

Starting with the original function:

y = 2/(x + 1)

To find the inverse, swap x and y:

x = 2/(y + 1)

Now, solve for y.

Multiply both sides by (y + 1):

x(y + 1) = 2

Expanding the left side:

xy + x = 2

Now, subtract x from both sides:

xy = 2 - x

Then, divide both sides by x:

y = (2 - x)/x

Wait, that seems different from my previous answer. Wait, but hold on, if I have y = (2 - x)/x, then that would be the inverse function? Let me check.

Alternatively, maybe I made a mistake in algebra here. Let me redo that step.

Starting again from x = 2/(y + 1).

Multiply both sides by (y + 1):

x(y + 1) = 2

Which is:

xy + x = 2

Then, subtract x:

xy = 2 - x

Then, divide both sides by x:

y = (2 - x)/x

Alternatively, that can be written as y = 2/x - 1. Let me check that:

(2 - x)/x = 2/x - x/x = 2/x - 1. Yes, that's correct.

So the inverse function f⁻¹(x) is 2/x - 1.

Therefore, to find f⁻¹(1/5), substitute x = 1/5 into this inverse function:

f⁻¹(1/5) = 2/(1/5) - 1

Calculating 2 divided by 1/5 is 2 * 5 = 10. Then subtract 1: 10 - 1 = 9.

So that gives me 9 again. So both methods agree. Therefore, the answer is 9.

But let me think again if there's another way to approach this. Maybe by using function composition?

If f⁻¹(1/5) is the value of x such that f(x) = 1/5, which is exactly what I did first. So solving f(x) = 1/5 gives x = 9, so f⁻¹(1/5) = 9.

Alternatively, maybe I can check if f(f⁻¹(1/5)) = 1/5. Let me see. If f⁻¹(1/5) is 9, then f(9) should be 1/5, which we already checked.

Alternatively, if I take the inverse function f⁻¹(x) = 2/x - 1, then f⁻¹(1/5) = 2/(1/5) - 1 = 10 - 1 = 9.

So all methods lead to the same answer.

Wait, but just to make sure that I didn't make a mistake in finding the inverse function. Let me check with another value. Let's say, for example, if x = 1, then f(1) = 2/(1 + 1) = 1. So f(1) = 1. Then, according to the inverse function, f⁻¹(1) should be 1. Let me check with the inverse function: f⁻¹(1) = 2/1 - 1 = 2 - 1 = 1. Correct.

Another test: Let me take x = 0. Then f(0) = 2/(0 + 1) = 2. So f(0) = 2. Then f⁻¹(2) should be 0. Let me compute f⁻¹(2) using the inverse function: 2/2 - 1 = 1 - 1 = 0. Correct.

Another test: Let me take x = 2. Then f(2) = 2/(2 + 1) = 2/3. Therefore, f⁻¹(2/3) should be 2. Let me compute f⁻¹(2/3): 2/(2/3) - 1 = 3 - 1 = 2. Correct.

So the inverse function seems to be working. Therefore, my calculation for f⁻¹(1/5) = 9 is correct.

Alternatively, maybe I can think about the graph of the function. The original function f(x) = 2/(x + 1) is a hyperbola with a vertical asymptote at x = -1 and a horizontal asymptote at y = 0. The inverse function would be a reflection over the line y = x, so it should also be a hyperbola.

But maybe that's overcomplicating. Since all my algebra checks out and the test cases work, I think I can be confident that the answer is 9.

Wait, but just to make sure, let me try solving the equation again.

Original equation: 2/(x + 1) = 1/5.

Cross-multiplying: 2 * 5 = 1 * (x + 1) => 10 = x + 1 => x = 9.

Yes, that's straightforward.

Alternatively, if I wanted to solve for x in the equation 2/(x + 1) = 1/5, I could also take reciprocals of both sides.

Taking reciprocals: (x + 1)/2 = 5/1 => x + 1 = 10 => x = 9.

Same result.

So, regardless of the method, I get x = 9.

Therefore, I think the answer is definitely 9.

I don't see any mistakes in my reasoning. All steps check out, and multiple methods lead to the same conclusion.

**Final Answer**
The value of $ f^{-1}\left(\frac{1}{5}\right) $ is \boxed{9}.

To find the value of $ f^{-1}\left(\frac{1}{5}\right) $, we begin by understanding what this notation means. The inverse function $ f^{-1}(y) $ gives the value of $ x $ such that $ f(x) = y $. In this case, we are looking for the $ x $ such that:

$$
f(x) = \frac{1}{5}
$$

Given the function:

$$
f(x) = \frac{2}{x + 1}
$$

We set this equal to $ \frac{1}{5} $ and solve for $ x $:

$$
\frac{2}{x + 1} = \frac{1}{5}
$$

---

### Step 1: Cross-Multiply

To eliminate the fractions, we cross-multiply:

$$
2 \cdot 5 = 1 \cdot (x + 1)
$$

$$
10 = x + 1
$$

---

### Step 2: Solve for $ x $

Subtract 1 from both sides:

$$
x = 10 - 1 = 9
$$

---

### Step 3: Verify the Result

To ensure the solution is correct, we substitute $ x = 9 $ back into the original function:

$$
f(9) = \frac{2}{9 + 1} = \frac{2}{10} = \frac{1}{5}
$$

This confirms that $ f(9) = \frac{1}{5} $, so $ f^{-1}\left(\frac{1}{5}\right) = 9 $.

---

### Alternative Method: Finding the Inverse Function

We can also find the inverse function explicitly by switching $ x $ and $ y $ in the original function:

$$
y = \frac{2}{x + 1}
$$

Swap $ x $ and $ y $:

$$
x = \frac{2}{y + 1}
$$

Now solve for $ y $:

$$
x(y + 1) = 2 \Rightarrow xy + x = 2 \Rightarrow xy = 2 - x \Rightarrow y = \frac{2 - x}{x}
$$

So the inverse function is:

$$
f^{-1}(x) = \frac{2 - x}{x}
$$

Now evaluate $ f^{-1}\left(\frac{1}{5}\right) $:

$$
f^{-1}\left(\frac{1}{5}\right) = \frac{2 - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{10 - 1}{5}}{\frac{1}{5}} = \frac{9}{5} \div \frac{1}{5} = 9
$$

---

### Final Answer

$$
\boxed{9}
$$

Solution: $f^{-1}\left(\frac{1}{5}\right)$ is defined as the number $x$ such that $f(x)=\frac{1}{5}$. Thus, we solve the equation $$\frac{2}{x+1} = \frac{1}{5}.$$Multiplying both sides by $5(x+1)$, we have $$10 = x+1.$$Subtracting $1$ from both sides gives $x=\boxed{9}$.


Explanation:
Okay, so I need to find the distance between the foci of a hyperbola. The problem gives me the equations of the asymptotes and a point that the hyperbola passes through. Let me recall what I know about hyperbolas and their asymptotes.

First, hyperbolas have two asymptotes, which are straight lines that the hyperbola approaches but never touches. The equations of the asymptotes depend on the orientation of the hyperbola. If the hyperbola is horizontal, the asymptotes are of the form y = ±(b/a)x + k, and if it's vertical, they are y = ±(a/b)x + k, where (h, k) is the center of the hyperbola. Wait, actually, maybe I need to be more precise here.

The standard form of a hyperbola centered at (h, k) is either:

1. $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ for a horizontal hyperbola, with asymptotes y = ±(b/a)(x - h) + k.

2. $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ for a vertical hyperbola, with asymptotes y = ±(a/b)(x - h) + k.

But in this problem, the asymptotes are given as y = 2x - 3 and y = 17 - 2x. Let me note that these two lines have slopes 2 and -2. Since the slopes are positive and negative, they are symmetric with respect to the x-axis and y-axis? Wait, not necessarily. Let me think.

The slopes of the asymptotes for a hyperbola are determined by the ratio of a and b. For a horizontal hyperbola, the slopes are ±(b/a), and for a vertical hyperbola, they are ±(a/b). So if the slopes here are 2 and -2, then depending on the orientation, we can figure out which one it is.

But first, let me find the center of the hyperbola. The center is the intersection point of the asymptotes. So I need to solve the system of equations:

y = 2x - 3

and

y = 17 - 2x.

Let me set them equal to each other:

2x - 3 = 17 - 2x

Adding 2x to both sides:

4x - 3 = 17

Adding 3 to both sides:

4x = 20

Divide by 4:

x = 5

Then plug back into one of the equations to find y. Let's use the first one:

y = 2(5) - 3 = 10 - 3 = 7

So the center of the hyperbola is at (5, 7). Wait, that's interesting because the hyperbola passes through the point (4, 7). So the center is (5,7), and the point (4,7) is one unit to the left of the center. Hmm, maybe that will be useful later.

Now, knowing the center, I can write the equations of the asymptotes in terms of the center. The asymptotes are y = 2x - 3 and y = -2x + 17. Let me rewrite them in terms of the center (5,7). Let me check:

For the first asymptote, y = 2x - 3. Let me write it as y = 2(x - 5) + 7? Let me check:

2(x - 5) + 7 = 2x - 10 + 7 = 2x - 3. Yes! Perfect. Similarly, the second asymptote is y = -2x + 17. Let me write that as y = -2(x - 5) + 7? Let's check:

-2(x - 5) + 7 = -2x + 10 + 7 = -2x + 17. Yes! Perfect. So the asymptotes are y = ±2(x - 5) + 7. Therefore, the slopes are ±2, which means that the hyperbola has slopes of asymptotes ±(b/a) or ±(a/b) depending on orientation.

So, since the slopes are ±2, which is a positive number, and since the slopes for horizontal hyperbola are ±(b/a), and for vertical hyperbola are ±(a/b). So which one is it?

If the hyperbola is horizontal, then the slopes are ±(b/a). If vertical, ±(a/b). So if the slopes are ±2, then depending on which is larger, a or b. But we don't know yet. However, since the center is (5,7), and the hyperbola passes through (4,7), which is directly to the left of the center. Let me think about that.

If the hyperbola is horizontal, then the transverse axis is horizontal, so the vertices are left and right from the center. If the hyperbola is vertical, the transverse axis is vertical, so the vertices are up and down from the center. Since the point (4,7) is horizontally aligned with the center (since y-coordinate is same), if the hyperbola is horizontal, then (4,7) could be a vertex or a point on the hyperbola. If it's vertical, then (4,7) would be a point on the hyperbola but not necessarily a vertex.

But maybe I can figure out the orientation based on the slopes. Let me think. Let me suppose that the hyperbola is horizontal. Then the slopes of the asymptotes are ±(b/a) = ±2. So b/a = 2, so b = 2a. Alternatively, if it's vertical, then slopes are ±(a/b) = ±2, so a/b = 2, so a = 2b. So depending on which case, we can have different relations between a and b.

But how do I know which one it is? Maybe I need to use the point (4,7) to figure out which case it is. Let me try both possibilities.

First, let's assume it's a horizontal hyperbola. Then the standard equation is:

$\frac{(x - 5)^2}{a^2} - \frac{(y - 7)^2}{b^2} = 1$

With asymptotes y = ±(b/a)(x - 5) + 7. Which we know are y = ±2(x - 5) + 7. Therefore, b/a = 2, so b = 2a.

So then, the equation becomes:

$\frac{(x - 5)^2}{a^2} - \frac{(y - 7)^2}{(2a)^2} = 1$

Simplify:

$\frac{(x - 5)^2}{a^2} - \frac{(y - 7)^2}{4a^2} = 1$

Now, since the hyperbola passes through (4,7), let's plug that point into the equation.

Plugging x = 4, y = 7:

$\frac{(4 - 5)^2}{a^2} - \frac{(7 - 7)^2}{4a^2} = 1$

Simplify:

$\frac{(-1)^2}{a^2} - 0 = 1$

Which is:

$\frac{1}{a^2} = 1$

Therefore, a² = 1, so a = 1 (since a is positive). Then, since b = 2a, b = 2*1 = 2. Therefore, b² = 4.

So the equation of the hyperbola would be:

$\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$

Let me check if this hyperbola actually passes through (4,7). Plugging in, we get ( (4-5)^2 ) /1 - ( (7-7)^2 ) /4 = 1 - 0 = 1. Yes, that works. So that seems okay.

But let me check if this hyperbola has the correct asymptotes. The asymptotes for this hyperbola would be y - 7 = ±(b/a)(x - 5) => y - 7 = ±2(x - 5). Which gives y = 2x - 10 + 7 = 2x - 3 and y = -2x + 10 + 7 = -2x + 17. Which matches the given asymptotes. Perfect. So that seems to check out.

So if this is the case, then the hyperbola is horizontal, with a = 1, b = 2. Then, the distance between the foci is 2c, where c² = a² + b². So c² = 1 + 4 = 5, so c = √5. Therefore, the distance between the foci is 2√5. So is that the answer? Wait, but let me check if there's another possibility where the hyperbola is vertical.

Let me try assuming the hyperbola is vertical. Then the standard equation is:

$\frac{(y - 7)^2}{a^2} - \frac{(x - 5)^2}{b^2} = 1$

With asymptotes y - 7 = ±(a/b)(x - 5). Given that the asymptotes are y = ±2(x - 5) + 7, so the slopes are ±2. Therefore, (a/b) = 2, so a = 2b.

So then, the equation becomes:

$\frac{(y - 7)^2}{(2b)^2} - \frac{(x - 5)^2}{b^2} = 1$

Simplify:

$\frac{(y - 7)^2}{4b^2} - \frac{(x - 5)^2}{b^2} = 1$

Now, plug in the point (4,7):

$\frac{(7 - 7)^2}{4b^2} - \frac{(4 - 5)^2}{b^2} = 1$

Simplify:

0 - $\frac{(-1)^2}{b^2}$ = 1 => -1/b² = 1 => -1 = b². Which is impossible because b² can't be negative. Therefore, this case is invalid. Therefore, the hyperbola cannot be vertical. Therefore, the hyperbola must be horizontal, with a = 1, b = 2, and c² = a² + b² = 1 + 4 = 5, so c = √5, so distance between foci is 2c = 2√5. Therefore, the answer is 2√5. Wait, but let me check again.

But hold on, the problem says "the hyperbola passes through the point (4,7)". In the horizontal case, we found that (4,7) is on the hyperbola. But in the vertical case, we get an impossible equation. Therefore, the hyperbola must be horizontal. Therefore, the distance between the foci is 2√5. Therefore, the answer is 2√5. But let me check again if I did everything correctly.

Wait, but let me think again. The hyperbola is horizontal, with center (5,7), a = 1, so the vertices are at (5 ± a, 7) which is (6,7) and (4,7). Wait, so the point (4,7) is actually one of the vertices? Because the vertex is at (5 - 1, 7) = (4,7). So that point is a vertex. So that makes sense. So the hyperbola has vertices at (4,7) and (6,7). Therefore, the distance between the foci is 2c, where c is sqrt(a² + b²) = sqrt(1 + 4) = sqrt(5). Therefore, the distance between foci is 2*sqrt(5). Therefore, the answer is 2√5. So that seems correct.

But let me check if there's another way to approach this problem, just to confirm.

Alternatively, since we know the center is (5,7), and the slopes of the asymptotes are ±2, then for a hyperbola, the slopes are ±(b/a) if horizontal, or ±(a/b) if vertical. Since we saw that vertical case leads to inconsistency, so horizontal case is correct. Then, since the hyperbola passes through (4,7), which is a vertex, so that gives us a = 1. Then, since the slopes are ±(b/a) = ±2, so b/a = 2, so b = 2. Then c^2 = a^2 + b^2 = 1 + 4 = 5, so c = sqrt(5), so distance between foci is 2c = 2*sqrt(5). So that's the same answer.

Alternatively, maybe I can write the equation of the hyperbola in general form and use the point to find the parameters. Let me try that approach.

The general equation of a hyperbola with asymptotes y = 2x - 3 and y = -2x + 17 can be written as (y - 2x + 3)(y + 2x - 17) = k, where k is a constant. Because the asymptotes are the lines that the hyperbola approaches, so the equation of the hyperbola can be written as the product of the equations of the asymptotes equals some constant. Let me check that.

The asymptotes are y = 2x - 3 and y = -2x + 17. Let me rearrange them to standard form:

First asymptote: y - 2x + 3 = 0

Second asymptote: y + 2x - 17 = 0

Therefore, the product of these two equations is (y - 2x + 3)(y + 2x - 17) = k. So expanding this product:

Let me compute (y - 2x + 3)(y + 2x - 17). Let me denote terms:

Let me write it as [(y + 3) - 2x][(y - 17) + 2x]. Wait, maybe expanding directly:

Multiply term by term:

First, y*(y) = y²

y*(2x) = 2xy

y*(-17) = -17y

(-2x)*(y) = -2xy

(-2x)*(2x) = -4x²

(-2x)*(-17) = 34x

3*(y) = 3y

3*(2x) = 6x

3*(-17) = -51

So combining all terms:

y² + 2xy -17y -2xy -4x² +34x +3y +6x -51

Now, let's combine like terms:

- The y² term: y²

- The xy terms: 2xy - 2xy = 0

- The y terms: -17y + 3y = -14y

- The x² term: -4x²

- The x terms: 34x + 6x = 40x

- The constant term: -51

So overall, the product is:

y² - 4x² -14y + 40x -51 = k

Therefore, the equation of the hyperbola is:

y² - 4x² -14y + 40x -51 = k

But we can rearrange this equation to standard form. However, since we know that the hyperbola passes through (4,7), we can substitute x=4, y=7 into this equation to find k.

Let me do that:

Left side: (7)^2 - 4*(4)^2 -14*(7) + 40*(4) -51

Calculate each term:

7² = 49

-4*(16) = -64

-14*7 = -98

40*4 = 160

-51 remains.

So total:

49 - 64 - 98 + 160 -51

Let me compute step by step:

Start with 49 - 64 = -15

-15 - 98 = -113

-113 + 160 = 47

47 - 51 = -4

Therefore, k = -4. Therefore, the equation of the hyperbola is:

y² - 4x² -14y + 40x -51 = -4

Which simplifies to:

y² - 4x² -14y + 40x -51 + 4 = 0 => y² - 4x² -14y + 40x -47 = 0

But maybe it's better to rearrange terms and complete the squares to get into standard form.

Let me group the y terms and x terms:

(y² -14y) -4x² + 40x = 47

Let me complete the square for y and x.

Starting with y² -14y:

Take half of -14, which is -7, square it: 49. So add and subtract 49:

(y² -14y + 49 - 49) = (y - 7)^2 - 49

Similarly for the x terms: -4x² + 40x. Let me factor out the coefficient of x² first:

-4(x² - 10x). Now complete the square inside the parentheses:

Take half of -10, which is -5, square it: 25. So add and subtract 25:

-4[(x² -10x +25 -25)] = -4[(x - 5)^2 -25] = -4(x - 5)^2 + 100

So putting it all back into the equation:

(y - 7)^2 - 49 -4(x - 5)^2 + 100 = 47

Simplify constants:

-49 + 100 = 51

So:

(y - 7)^2 -4(x - 5)^2 + 51 = 47

Subtract 51 from both sides:

(y - 7)^2 -4(x - 5)^2 = 47 - 51 = -4

Multiply both sides by -1:

4(x - 5)^2 - (y - 7)^2 = 4

Divide both sides by 4 to get 1 on the right:

$\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$

Which is exactly the same equation as before! So that confirms that the standard form is $\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$, so a² = 1, b² = 4, so c² = a² + b² = 5, so c = √5, so distance between foci is 2c = 2√5. Therefore, the answer is 2√5.

Therefore, I think this is correct. Let me check once more if there's any possible mistake.

Wait, in the general equation approach, we found that the hyperbola equation is $\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$, which is a horizontal hyperbola with center (5,7), a = 1, b = 2. The foci are located at (h ± c, k), so (5 ± √5, 7). Therefore, the distance between them is 2√5. So that's correct.

Alternatively, if I had made a mistake in assuming the hyperbola is horizontal, but since the vertical case led to an inconsistency, that's not possible. Therefore, I think this is solid.

Therefore, the answer is 2√5. So I need to write that in boxed notation.

**Final Answer**
\boxed{2\sqrt{5}}

To find the **distance between the foci** of the hyperbola, we are given:

- The **asymptotes**:
  $ y = 2x - 3 $ and $ y = 17 - 2x $

- A **point on the hyperbola**: $ (4, 7) $

---

### Step 1: Find the Center of the Hyperbola

The **center** of a hyperbola is the **intersection point** of its asymptotes. So, we solve the system:

$$
\begin{cases}
y = 2x - 3 \\
y = 17 - 2x
\end{cases}
$$

Set the two expressions for $ y $ equal:

$$
2x - 3 = 17 - 2x \Rightarrow 4x = 20 \Rightarrow x = 5
$$

Substitute $ x = 5 $ into either equation to find $ y $:

$$
y = 2(5) - 3 = 10 - 3 = 7
$$

So, the **center** of the hyperbola is at $ (5, 7) $.

---

### Step 2: Determine the Orientation and Asymptote Slopes

The asymptotes are:

- $ y = 2x - 3 $
- $ y = -2x + 17 $

Rewriting them in point-slope form with the center $ (5, 7) $:

- $ y = 2(x - 5) + 7 $
- $ y = -2(x - 5) + 7 $

So the asymptotes are:

$$
y = \pm 2(x - 5) + 7
$$

This indicates the **slopes** of the asymptotes are $ \pm 2 $, which is consistent with a **horizontal hyperbola**. For a horizontal hyperbola, the slopes of the asymptotes are $ \pm \frac{b}{a} $, so:

$$
\frac{b}{a} = 2 \Rightarrow b = 2a
$$

---

### Step 3: Use the Given Point to Find $ a $ and $ b $

We are told the hyperbola passes through the point $ (4, 7) $. Since the center is $ (5, 7) $, this point lies directly to the left of the center, at a distance of 1 unit. This suggests that $ (4, 7) $ is a **vertex** of the hyperbola.

For a horizontal hyperbola, the vertices are located at $ (h \pm a, k) $. Since the center is $ (5, 7) $, and the vertex is at $ (4, 7) $, we have:

$$
a = 1
$$

Then, since $ b = 2a $, we get:

$$
b = 2
$$

---

### Step 4: Find the Distance Between the Foci

For a hyperbola, the distance between the foci is $ 2c $, where:

$$
c = \sqrt{a^2 + b^2}
$$

Substitute $ a = 1 $, $ b = 2 $:

$$
c = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
$$

So the **distance between the foci** is:

$$
2c = 2\sqrt{5}
$$

---

### Final Answer

$$
\boxed{2\sqrt{5}}
$$

[Assistant]
Explanation:
Okay, so I need to find the distance between the foci of a hyperbola. The problem gives me the equations of the asymptotes and a point that the hyperbola passes through. Let me recall what I know about hyperbolas and their asymptotes.

First, hyperbolas have two asymptotes, which are straight lines that the hyperbola approaches but never touches. The equations of the asymptotes depend on the orientation of the hyperbola. If the hyperbola is horizontal, the asymptotes are of the form y = ±(b/a)x + k, and if it's vertical, they are y = ±(a/b)x + k, where (h, k) is the center of the hyperbola. Wait, actually, maybe I need to be more precise here.

The standard form of a hyperbola centered at (h, k) is either:

1. $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ for a horizontal hyperbola, with asymptotes y = ±(b/a)(x - h) + k.

2. $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ for a vertical hyperbola, with asymptotes y = ±(a/b)(x - h) + k.

But in this problem, the asymptotes are given as y = 2x - 3 and y = 17 - 2x. Let me note that these two lines have slopes 2 and -2. Since the slopes are positive and negative, they are symmetric with respect to the x-axis and y-axis? Wait, not necessarily. Let me think.

The slopes of the asymptotes for a hyperbola are determined by the ratio of a and b. For a horizontal hyperbola, the slopes are ±(b/a), and for a vertical hyperbola, they are ±(a/b). So if the slopes here are 2 and -2, then depending on the orientation, we can figure out which one it is.

But first, let me find the center of the hyperbola. The center is the intersection point of the asymptotes. So I need to solve the system of equations:

y = 2x - 3

and

y = 17 - 2x.

Let me set them equal to each other:

2x - 3 = 17 - 2x

Adding 2x to both sides:

4x - 3 = 17

Adding 3 to both sides:

4x = 20

Divide by 4:

x = 5

Then plug back into one of the equations to find y. Let's use the first one:

y = 2(5) - 3 = 10 - 3 = 7

So the center of the hyperbola is at (5, 7). Wait, that's interesting because the hyperbola passes through the point (4, 7). So the center is (5,7), and the point (4,7) is one unit to the left of the center. Hmm, maybe that will be useful later.

Now, knowing the center, I can write the equations of the asymptotes in terms of the center. The asymptotes are y = 2x - 3 and y = -2x + 17. Let me rewrite them in terms of the center (5,7). Let me check:

For the first asymptote, y = 2x - 3. Let me write it as y = 2(x - 5) + 7? Let me check:

2(x - 5) + 7 = 2x - 10 + 7 = 2x - 3. Yes! Perfect. Similarly, the second asymptote is y = -2x + 17. Let me write that as y = -2(x - 5) + 7? Let's check:

-2(x - 5) + 7 = -2x + 10 + 7 = -2x + 17. Yes! Perfect. So the asymptotes are y = ±2(x - 5) + 7. Therefore, the slopes are ±2, which means that the hyperbola has slopes of asymptotes ±(b/a) or ±(a/b) depending on orientation.

So, since the slopes are ±2, which is a positive number, and since the slopes for horizontal hyperbola are ±(b/a), and for vertical hyperbola are ±(a/b). So which one is it?

If the hyperbola is horizontal, then the slopes are ±(b/a). If vertical, ±(a/b). So if the slopes are ±2, then depending on which is larger, a or b. But we don't know yet. However, since the center is (5,7), and the hyperbola passes through (4,7), which is directly to the left of the center. Let me think about that.

If the hyperbola is horizontal, then the transverse axis is horizontal, so the vertices are left and right from the center. If the hyperbola is vertical, the transverse axis is vertical, so the vertices are up and down from the center. Since the point (4,7) is horizontally aligned with the center (since y-coordinate is same), if the hyperbola is horizontal, then (4,7) could be a vertex or a point on the hyperbola. If it's vertical, then (4,7) would be a point on the hyperbola but not necessarily a vertex.

But maybe I can figure out the orientation based on the slopes. Let me think. Let me suppose that the hyperbola is horizontal. Then the slopes of the asymptotes are ±(b/a) = ±2. So b/a = 2, so b = 2a. Alternatively, if it's vertical, then slopes are ±(a/b) = ±2, so a/b = 2, so a = 2b. So depending on which case, we can have different relations between a and b.

But how do I know which one it is? Maybe I need to use the point (4,7) to figure out which case it is. Let me try both possibilities.

First, let's assume it's a horizontal hyperbola. Then the standard equation is:

$\frac{(x - 5)^2}{a^2} - \frac{(y - 7)^2}{b^2} = 1$

With asymptotes y = ±(b/a)(x - 5) + 7. Which we know are y = ±2(x - 5) + 7. Therefore, b/a = 2, so b = 2a.

So then, the equation becomes:

$\frac{(x - 5)^2}{a^2} - \frac{(y - 7)^2}{(2a)^2} = 1$

Simplify:

$\frac{(x - 5)^2}{a^2} - \frac{(y - 7)^2}{4a^2} = 1$

Now, since the hyperbola passes through (4,7), let's plug that point into the equation.

Plugging x = 4, y = 7:

$\frac{(4 - 5)^2}{a^2} - \frac{(7 - 7)^2}{4a^2} = 1$

Simplify:

$\frac{(-1)^2}{a^2} - 0 = 1$

Which is:

$\frac{1}{a^2} = 1$

Therefore, a² = 1, so a = 1 (since a is positive). Then, since b = 2a, b = 2*1 = 2. Therefore, b² = 4.

So the equation of the hyperbola would be:

$\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$

Let me check if this hyperbola actually passes through (4,7). Plugging in, we get ( (4-5)^2 ) /1 - ( (7-7)^2 ) /4 = 1 - 0 = 1. Yes, that works. So that seems okay.

But let me check if this hyperbola has the correct asymptotes. The asymptotes for this hyperbola would be y - 7 = ±(b/a)(x - 5) => y - 7 = ±2(x - 5). Which gives y = 2x - 10 + 7 = 2x - 3 and y = -2x + 10 + 7 = -2x + 17. Which matches the given asymptotes. Perfect. So that seems to check out.

So if this is the case, then the hyperbola is horizontal, with a = 1, b = 2. Then, the distance between the foci is 2c, where c² = a² + b². So c² = 1 + 4 = 5, so c = √5. Therefore, the distance between the foci is 2√5. So is that the answer? Wait, but let me check if there's another possibility where the hyperbola is vertical.

Let me try assuming the hyperbola is vertical. Then the standard equation is:

$\frac{(y - 7)^2}{a^2} - \frac{(x - 5)^2}{b^2} = 1$

With asymptotes y - 7 = ±(a/b)(x - 5). Given that the asymptotes are y = ±2(x - 5) + 7, so the slopes are ±2. Therefore, (a/b) = 2, so a = 2b.

So then, the equation becomes:

$\frac{(y - 7)^2}{(2b)^2} - \frac{(x - 5)^2}{b^2} = 1$

Simplify:

$\frac{(y - 7)^2}{4b^2} - \frac{(x - 5)^2}{b^2} = 1$

Now, plug in the point (4,7):

$\frac{(7 - 7)^2}{4b^2} - \frac{(4 - 5)^2}{b^2} = 1$

Simplify:

0 - $\frac{(-1)^2}{b^2}$ = 1 => -1/b² = 1 => -1 = b². Which is impossible because b² can't be negative. Therefore, this case is invalid. Therefore, the hyperbola cannot be vertical. Therefore, the hyperbola must be horizontal, with a = 1, b = 2, and c² = a² + b² = 1 + 4 = 5, so c = √5, so distance between foci is 2c = 2√5. Therefore, the answer is 2√5. Wait, but let me check again.

But hold on, the problem says "the hyperbola passes through the point (4,7)". In the horizontal case, we found that (4,7) is on the hyperbola. But in the vertical case, we get an impossible equation. Therefore, the hyperbola must be horizontal. Therefore, the distance between the foci is 2√5. Therefore, the answer is 2√5. But let me check again if I did everything correctly.

Wait, but let me think again. The hyperbola is horizontal, with center (5,7), a = 1, so the vertices are at (5 ± a, 7) which is (6,7) and (4,7). Wait, so the point (4,7) is actually one of the vertices? Because the vertex is at (5 - 1, 7) = (4,7). So that point is a vertex. So that makes sense. So the hyperbola has vertices at (4,7) and (6,7). Therefore, the distance between the foci is 2c, where c is sqrt(a² + b²) = sqrt(1 + 4) = sqrt(5). Therefore, the distance between foci is 2*sqrt(5). Therefore, the answer is 2√5. So that seems correct.

But let me check if there's another way to approach this problem, just to confirm.

Alternatively, since we know the center is (5,7), and the slopes of the asymptotes are ±2, then for a hyperbola, the slopes are ±(b/a) if horizontal, or ±(a/b) if vertical. Since we saw that vertical case leads to inconsistency, so horizontal case is correct. Then, since the hyperbola passes through (4,7), which is a vertex, so that gives us a = 1. Then, since the slopes are ±(b/a) = ±2, so b/a = 2, so b = 2. Then c^2 = a^2 + b^2 = 1 + 4 = 5, so c = sqrt(5), so distance between foci is 2c = 2*sqrt(5). So that's the same answer.

Alternatively, maybe I can write the equation of the hyperbola in general form and use the point to find the parameters. Let me try that approach.

The general equation of a hyperbola with asymptotes y = 2x - 3 and y = -2x + 17 can be written as (y - 2x + 3)(y + 2x - 17) = k, where k is a constant. Because the asymptotes are the lines that the hyperbola approaches, so the equation of the hyperbola can be written as the product of the equations of the asymptotes equals some constant. Let me check that.

The asymptotes are y = 2x - 3 and y = -2x + 17. Let me rearrange them to standard form:

First asymptote: y - 2x + 3 = 0

Second asymptote: y + 2x - 17 = 0

Therefore, the product of these two equations is (y - 2x + 3)(y + 2x - 17) = k. So expanding this product:

Let me compute (y - 2x + 3)(y + 2x - 17). Let me denote terms:

Let me write it as [(y + 3) - 2x][(y - 17) + 2x]. Wait, maybe expanding directly:

Multiply term by term:

First, y*(y) = y²

y*(2x) = 2xy

y*(-17) = -17y

(-2x)*(y) = -2xy

(-2x)*(2x) = -4x²

(-2x)*(-17) = 34x

3*(y) = 3y

3*(2x) = 6x

3*(-17) = -51

So combining all terms:

y² + 2xy -17y -2xy -4x² +34x +3y +6x -51

Now, let's combine like terms:

- The y² term: y²

- The xy terms: 2xy - 2xy = 0

- The y terms: -17y + 3y = -14y

- The x² term: -4x²

- The x terms: 34x + 6x = 40x

- The constant term: -51

So overall, the product is:

y² - 4x² -14y + 40x -51 = k

Therefore, the equation of the hyperbola is:

y² - 4x² -14y + 40x -51 = k

But we can rearrange this equation to standard form. However, since we know that the hyperbola passes through (4,7), we can substitute x=4, y=7 into this equation to find k.

Let me do that:

Left side: (7)^2 - 4*(4)^2 -14*(7) + 40*(4) -51

Calculate each term:

7² = 49

-4*(16) = -64

-14*7 = -98

40*4 = 160

-51 remains.

So total:

49 - 64 - 98 + 160 -51

Let me compute step by step:

Start with 49 - 64 = -15

-15 - 98 = -113

-113 + 160 = 47

47 - 51 = -4

Therefore, k = -4. Therefore, the equation of the hyperbola is:

y² - 4x² -14y + 40x -51 = -4

Which simplifies to:

y² - 4x² -14y + 40x -51 + 4 = 0 => y² - 4x² -14y + 40x -47 = 0

But maybe it's better to rearrange terms and complete the squares to get into standard form.

Let me group the y terms and x terms:

(y² -14y) -4x² + 40x = 47

Let me complete the square for y and x.

Starting with y² -14y:

Take half of -14, which is -7, square it: 49. So add and subtract 49:

(y² -14y + 49 - 49) = (y - 7)^2 - 49

Similarly for the x terms: -4x² + 40x. Let me factor out the coefficient of x² first:

-4(x² - 10x). Now complete the square inside the parentheses:

Take half of -10, which is -5, square it: 25. So add and subtract 25:

-4[(x² -10x +25 -25)] = -4[(x - 5)^2 -25] = -4(x - 5)^2 + 100

So putting it all back into the equation:

(y - 7)^2 - 49 -4(x - 5)^2 + 100 = 47

Simplify constants:

-49 + 100 = 51

So:

(y - 7)^2 -4(x - 5)^2 + 51 = 47

Subtract 51 from both sides:

(y - 7)^2 -4(x - 5)^2 = 47 - 51 = -4

Multiply both sides by -1:

4(x - 5)^2 - (y - 7)^2 = 4

Divide both sides by 4 to get 1 on the right:

$\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$

Which is exactly the same equation as before! So that confirms that the standard form is $\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$, so a² = 1, b² = 4, so c² = a² + b² = 5, so c = √5, so distance between foci is 2c = 2√5. Therefore, the answer is 2√5.

Therefore, I think this is correct. Let me check once more if there's any possible mistake.

Wait, in the general equation approach, we found that the hyperbola equation is $\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1$, which is a horizontal hyperbola with center (5,7), a = 1, b = 2. The foci are located at (h ± c, k), so (5 ± √5, 7). Therefore, the distance between them is 2√5. So that's correct.

Alternatively, if I had made a mistake in assuming the hyperbola is horizontal, but since the vertical case led to an inconsistency, that's not possible. Therefore, I think this is solid.

Therefore, the answer is 2√5. So I need to write that in boxed notation.

**Final Answer**
\boxed{2\sqrt{5}}

To find the **distance between the foci** of the hyperbola, we are given:

- The **asymptotes**:
  $ y = 2x - 3 $ and $ y = 17 - 2x $

- A **point on the hyperbola**: $ (4, 7) $

---

### Step 1: Find the Center of the Hyperbola

The **center** of a hyperbola is the **intersection point** of its asymptotes. So, we solve the system:

$$
\begin{cases}
y = 2x - 3 \\
y = 17 - 2x
\end{cases}
$$

Set the two expressions for $ y $ equal:

$$
2x - 3 = 17 - 2x \Rightarrow 4x = 20 \Rightarrow x = 5
$$

Substitute $ x = 5 $ into either equation to find $ y $:

$$
y = 2(5) - 3 = 10 - 3 = 7
$$

So, the **center** of the hyperbola is at $ (5, 7) $.

---

### Step 2: Determine the Orientation and Asymptote Slopes

The asymptotes are:

- $ y = 2x - 3 $
- $ y = -2x + 17 $

Rewriting them in point-slope form with the center $ (5, 7) $:

- $ y = 2(x - 5) + 7 $
- $ y = -2(x - 5) + 7 $

So the asymptotes are:

$$
y = \pm 2(x - 5) + 7
$$

This indicates the **slopes** of the asymptotes are $ \pm 2 $, which is consistent with a **horizontal hyperbola**. For a horizontal hyperbola, the slopes of the asymptotes are $ \pm \frac{b}{a} $, so:

$$
\frac{b}{a} = 2 \Rightarrow b = 2a
$$

---

### Step 3: Use the Given Point to Find $ a $ and $ b $

We are told the hyperbola passes through the point $ (4, 7) $. Since the center is $ (5, 7) $, this point lies directly to the left of the center, at a distance of 1 unit. This suggests that $ (4, 7) $ is a **vertex** of the hyperbola.

For a horizontal hyperbola, the vertices are located at $ (h \pm a, k) $. Since the center is $ (5, 7) $, and the vertex is at $ (4, 7) $, we have:

$$
a = 1
$$

Then, since $ b = 2a $, we get:

$$
b = 2
$$

---

### Step 4: Find the Distance Between the Foci

For a hyperbola, the distance between the foci is $ 2c $, where:

$$
c = \sqrt{a^2 + b^2}
$$

Substitute $ a = 1 $, $ b = 2 $:

$$
c = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
$$

So the **distance between the foci** is:

$$
2c = 2\sqrt{5}
$$

---

### Final Answer

$$
\boxed{2\sqrt{5}}
$$

Solution:
The intersection of the asymptotes is $(5,7),$ so this is the center of the hyperbola.  Since the slopes of the asymptotes are $\pm 2,$ the equation of the hyperbola can be written in the form
\[(x - 5)^2 - \frac{(y - 7)^2}{4} = d\]for some constant $d.$  Setting $x = 4$ and $y = 7,$ we get $d = 1,$ so the equation is
\[\frac{(x - 5)^2}{1} - \frac{(y - 7)^2}{4} = 1.\]Then $a^2 = 1$ and $b^2 = 4,$ so $c^2 = a^2 + b^2 = 5,$ which means $c = \sqrt{5}.$  Therefore, the distance between the foci is $2c = \boxed{2 \sqrt{5}}.$