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, but in base 7, digits can be from 0 to 6, and in base 8, digits can be from 0 to 7. However, the digits in the numbers are only those that are present in the representations.

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, the digits in base 7 are 1, 5, 5, 6.

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

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

But let me check if there is any other digit. For example, in base 7, digits are 0-6, but the number 629 in base 7 is 1556, so digits are 1,5,5,6. So digits used are 1,5,6. In base 8, digits are 0-7, but the number is 1165, so digits are 1,1,6,5. So digits used are 1,5,6. Therefore, same 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 number of digits that are same in both representations, considering their positions? For example, if a digit is in the same position in both numbers, 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 their positions.

Therefore, the answer is 3.

But let me check again. Maybe the problem is considering digits that are same in both representations, but considering the count. For example, if a digit appears multiple times in both representations, how is it counted?

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

Therefore, in our case, since all three digits (1,5,6) are common, the answer is 3.

But let me check again. Wait, maybe I need to check if there are digits that are same in both representations but with different values? Wait, no. The digits in base 7 and base 8 are digits in their respective bases. 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 in the representations.

But in our case, the digits in base 7 are 1,5,6. In base 8, digits are 1,5,6. So all three digits are common. Therefore, the answer is 3.

But let me check if there is a possibility that a digit in base 7 is same as a digit in base 8, but in different values? For example, if a digit is 5 in base 7 and 5 in base 8, but since 5 is a valid digit in both bases, it's the same digit.

Yes, digits are just symbols, so 5 is 5 regardless of the base. Therefore, the digits are same if they have the same symbol.

Therefore, the answer is 3.

But wait, let me check again. Maybe the problem is in the way digits are represented. For example, in base 7, the digits are 0-6, and in base 8, digits are 0-7. However, the digits in the numbers are only those that are present.

But in our case, the digits in base 7 are 1,5,6. In base 8, digits are 1,5,6. So all three digits are same. Therefore, answer is 3.

But maybe the problem is considering digits that are same in value? For example, if a digit in base 7 is equal to a digit in base 8. But since digits are symbols, they are same if they have the same symbol.

But maybe the problem is considering the numeric value? For example, if a digit in base 7 has the same numeric value as a digit in base 8. But that would be redundant, since digits in base 7 can't have values above 6, and digits in base 8 can have values up to 7. However, the digits present in the numbers are 1,5,6 in both. So their numeric values are same.

But regardless, the problem says "same digits", which would refer to the symbols. For example, the digit '1' is same in both bases. So the answer is 3.

But let me check again. Maybe I made a mistake in the conversion. Let me check once more.

Base 7: 629 divided by 7:

First division: 629 /7 = 89.857... So 7*89=623, remainder 6. Correct.

Then 89 /7 = 12.714..., 7*12=84, remainder 5. Correct.

Then 12 /7 = 1.714..., 7*1=7, remainder 5. Correct.

Then 1 /7 = 0.142..., remainder 1. Correct.

So digits are 1,5,5,6.

Base 8: 629 /8 = 78.625, so 8*78=624, remainder 5. Correct.

Then 78 /8 = 9.75, 8*9=72, remainder 6. Correct.

Then 9 /8 = 1.125, remainder 1. Correct.

Then 1 /8 = 0.125, remainder 1. Correct.

Digits are 1,1,6,5.

So digits are correct.

Therefore, digits in base 7: {1,5,6}

Digits in base 8: {1,5,6}

Therefore, same digits: 1,5,6. So three digits.

But the example given in the problem says that 121_3 and 413_5 have one digit in common. Let me check that again.

121_3 is digits 1,2,1.

413_5 is digits 4,1,3.

Common digits: 1. So one digit.

Therefore, the answer is 1.

Therefore, in our problem, since there are three common digits, the answer is 3.

But wait, maybe the problem is considering digits that are same in both representations, but considering the count? For example, if a digit appears multiple times in both representations, how is it counted?

But the example says that even though 1 appears twice in the first number, it's only counted once. Therefore, the answer is the number of distinct digits.

Therefore, the answer is 3.

But let me check once more. Maybe the problem is in the way digits are represented. For example, in base 7, the digits are 1,5,5,6. In base 8, digits are 1,1,6,5. So if we list all digits in both representations, considering multiplicity, then:

Base 7 digits: 1,5,5,6

Base 8 digits: 1,1,6,5

So the digits that are common in both representations (with multiplicity) would be 1,5,6. But how many times?

But the problem says "how many of the same digits are found". The wording is ambiguous. However, given the example, it's about distinct digits.

Alternatively, 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, regardless of position.

But in that case, the answer would be the number of digits that are present in both representations.

But in the example, 121_3 and 413_5 have digits 1,2,1 and 4,1,3. The digits present in both are 1. So answer is 1.

Therefore, the answer is the number of distinct digits that are present in both representations.

Therefore, in our case, since both representations have digits 1,5,6, the answer is 3.

But let me check if there is any other digit. For example, in base 7, digits are 1,5,5,6. In base 8, digits are 1,1,6,5. So digits are 1,5,6.

Therefore, answer is 3.

But maybe the problem is considering digits that are same in both representations, but considering the digits in the numbers. For example, if a digit is present in both numbers, even if it's not in the same position.

But that's what we did.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, considering the digits in the same position.

But in that case, for our problem:

Base 7: 1 5 5 6

Base 8: 1 1 6 5

Comparing each 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, there are no digits in the same position. Therefore, the example contradicts that interpretation.

Therefore, the problem must be referring to digits that are present in both representations, regardless of their positions.

Therefore, the answer is 3.

But let me check once more. Maybe the problem is from a source where "same digits" refers to digits that are same in value, not symbols. For example, if a digit in base 7 is equal to a digit in base 8. But since digits are symbols, they are same if they have the same symbol.

But if we consider the numeric value, then digits in base 7 can be 0-6, and digits in base 8 can be 0-7. However, the digits present in the numbers are 1,5,6 in both. So their numeric values are same.

But again, the problem says "same digits", which would refer to the symbols.

Therefore, the answer is 3.

But maybe I need to check if there is a mistake in the problem's example. Let me check the example again.

The example says: 121_3 and 413_5 have one digit in common.

In base 3, 121 has digits 1, 2, 1. In base 5, 413 has digits 4, 1, 3. The common digit is 1. So only one digit.

Therefore, the answer is the number of distinct digits that are common.

Therefore, in our problem, since there are three common digits, the answer is 3.

But maybe the problem is from a source where digits are considered as numbers, not symbols. For example, if a digit in base 7 is 1, which is same as digit 1 in base 8. So same numeric value.

But in that case, the digits are same.

But in that case, the answer is still 3.

Therefore, I think the answer is 3.

But to be thorough, let me check if there is any other digit that is common.

In base 7 digits: 1,5,5,6.

In base 8 digits: 1,1,6,5.

So digits are 1,5,6.

No other digits.

Therefore, answer is 3.

But maybe the problem is in the way digits are counted. For example, if digits are considered with multiplicity.

For example, in base 7, there are two 5s and one 1 and one 6. In base 8, there are two 1s, one 6, and one 5.

So the digits that are same in both representations, considering multiplicity, would be:

For digit 1: appears once in base 7 and twice in base 8. So total same digits? If we count the minimum, it's 1.

For digit 5: appears twice in base 7 and once in base 8. So minimum is 1.

For digit 6: appears once in base 7 and once in base 8. So minimum is 1.

Total same digits: 1+1+1=3.

But if we count the maximum, it would be 2+1+1=4. But the example says that in the case of 121_3 and 413_5, which have digits 1,2,1 and 4,1,3, the common digits are 1. So if we count the minimum, it's 1.

Therefore, the answer is 3.

Therefore, I think the answer is 3.

But let me check once more.

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 numbers, regardless of how many times.

But that is the same as the number of distinct digits.

Therefore, answer is 3.

But maybe the problem is from a source where digits are considered as numbers, and the question is about how many digits (as numbers) are same in both representations.

But in that case, the digits are 1,5,6 in both, so three digits.

Therefore, answer is 3.

But maybe I need to check if there is a mistake in my conversion.

Wait, let me check base 7 again.

If I convert 1556 base 7 to decimal:

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

Base 8: 1165 base 8 is 1*8^3 + 1*8^2 + 6*8 + 5 = 512 + 64 + 48 + 5 = 512 + 64 is 576, 576 + 48 is 624, 624 + 5 is 629. Correct.

Therefore, the digits are correct.

Therefore, the answer is 3.

But maybe the problem is in the way digits are written. For example, in base 7, the digits are 1,5,5,6. In base 8, digits are 1,1,6,5. So if we list all digits in both representations, considering multiplicity, then:

Base 7 digits: [1,5,5,6]

Base 8 digits: [1,1,6,5]

The digits that are common are 1,5,6.

But if we count how many times they appear in both:

For digit 1: appears once in base 7 and twice in base 8.

For digit 5: appears twice in base 7 and once in base 8.

For digit 6: appears once in base 7 and once in base 8.

But the problem says "how many of the same digits are found". If it's asking for the total count of digits that are same in both representations, considering multiplicity, then it would be the sum of the minimum counts for each digit.

For digit 1: min(1,2)=1

For digit 5: min(2,1)=1

For digit 6: min(1,1)=1

Total: 1+1+1=3.

Alternatively, if it's asking for the total number of digits that are same in both representations, considering that each occurrence is counted, then it would be the sum of the minimum counts.

But the example given in the problem says that 121_3 and 413_5 have one digit in common. In that case, the digits in base 3 are [1,2,1], and in base 5 are [4,1,3]. The common digits are 1.

If we count the minimum counts: digit 1 appears twice in base 3 and once in base 5, so min(2,1)=1. Therefore, total is 1. Which matches the example.

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 interpretation.

Another possible interpretation: the problem is asking for the number of digits that are same in both representations, considering the digits in the same position.

For example, if a digit is same in the same position in both numbers, then it's counted.

In our case:

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 case, there are no digits in the same position. Therefore, this interpretation is invalid.

Therefore, the answer must be 3.

Therefore, I think the answer is 3.

But just to be thorough, let me check if there is any other digit that is common.

In base 7 digits: 1,5,5,6.

In base 8 digits: 1,1,6,5.

So digits are 1,5,6.

No other digits.

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 into 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, we get:

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

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, we get:

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

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.

---

### Step 4: Interpret the Question

The question asks:
> "How many of the same digits are found in the base 7 and base 8 representations of 629?"

This is interpreted as asking for the **number of distinct digits** that are **common** to both representations. Since the digits **1**, **5**, and **6** are all present in both, the answer is:

$$
\boxed{3}
$$

[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, but in base 7, digits can be from 0 to 6, and in base 8, digits can be from 0 to 7. However, the digits in the numbers are only those that are present in the representations.

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, the digits in base 7 are 1, 5, 5, 6.

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

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

But let me check if there is any other digit. For example, in base 7, digits are 0-6, but the number 629 in base 7 is 1556, so digits are 1,5,5,6. So digits used are 1,5,6. In base 8, digits are 0-7, but the number is 1165, so digits are 1,1,6,5. So digits used are 1,5,6. Therefore, same 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 number of digits that are same in both representations, considering their positions? For example, if a digit is in the same position in both numbers, 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 their positions.

Therefore, the answer is 3.

But let me check again. Maybe the problem is considering digits that are same in both representations, but considering the count. For example, if a digit appears multiple times in both representations, how is it counted?

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

Therefore, in our case, since all three digits (1,5,6) are common, the answer is 3.

But let me check again. Wait, maybe I need to check if there are digits that are same in both representations but with different values? Wait, no. The digits in base 7 and base 8 are digits in their respective bases. 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 in the representations.

But in our case, the digits in base 7 are 1,5,6. In base 8, digits are 1,5,6. So all three digits are common. Therefore, the answer is 3.

But let me check if there is a possibility that a digit in base 7 is same as a digit in base 8, but in different values? For example, if a digit is 5 in base 7 and 5 in base 8, but since 5 is a valid digit in both bases, it's the same digit.

Yes, digits are just symbols, so 5 is 5 regardless of the base. Therefore, the digits are same if they have the same symbol.

Therefore, the answer is 3.

But wait, let me check again. Maybe the problem is in the way digits are represented. For example, in base 7, the digits are 0-6, and in base 8, digits are 0-7. However, the digits in the numbers are only those that are present.

But in our case, the digits in base 7 are 1,5,6. In base 8, digits are 1,5,6. So all three digits are same. Therefore, answer is 3.

But maybe the problem is considering digits that are same in value? For example, if a digit in base 7 is equal to a digit in base 8. But since digits are symbols, they are same if they have the same symbol.

But maybe the problem is considering the numeric value? For example, if a digit in base 7 has the same numeric value as a digit in base 8. But that would be redundant, since digits in base 7 can't have values above 6, and digits in base 8 can have values up to 7. However, the digits present in the numbers are 1,5,6 in both. So their numeric values are same.

But regardless, the problem says "same digits", which would refer to the symbols. For example, the digit '1' is same in both bases. So the answer is 3.

But let me check again. Maybe I made a mistake in the conversion. Let me check once more.

Base 7: 629 divided by 7:

First division: 629 /7 = 89.857... So 7*89=623, remainder 6. Correct.

Then 89 /7 = 12.714..., 7*12=84, remainder 5. Correct.

Then 12 /7 = 1.714..., 7*1=7, remainder 5. Correct.

Then 1 /7 = 0.142..., remainder 1. Correct.

So digits are 1,5,5,6.

Base 8: 629 /8 = 78.625, so 8*78=624, remainder 5. Correct.

Then 78 /8 = 9.75, 8*9=72, remainder 6. Correct.

Then 9 /8 = 1.125, remainder 1. Correct.

Then 1 /8 = 0.125, remainder 1. Correct.

Digits are 1,1,6,5.

So digits are correct.

Therefore, digits in base 7: {1,5,6}

Digits in base 8: {1,5,6}

Therefore, same digits: 1,5,6. So three digits.

But the example given in the problem says that 121_3 and 413_5 have one digit in common. Let me check that again.

121_3 is digits 1,2,1.

413_5 is digits 4,1,3.

Common digits: 1. So one digit.

Therefore, the answer is 1.

Therefore, in our problem, since there are three common digits, the answer is 3.

But wait, maybe the problem is considering digits that are same in both representations, but considering the count? For example, if a digit appears multiple times in both representations, how is it counted?

But the example says that even though 1 appears twice in the first number, it's only counted once. Therefore, the answer is the number of distinct digits.

Therefore, the answer is 3.

But let me check once more. Maybe the problem is in the way digits are represented. For example, in base 7, the digits are 1,5,5,6. In base 8, digits are 1,1,6,5. So if we list all digits in both representations, considering multiplicity, then:

Base 7 digits: 1,5,5,6

Base 8 digits: 1,1,6,5

So the digits that are common in both representations (with multiplicity) would be 1,5,6. But how many times?

But the problem says "how many of the same digits are found". The wording is ambiguous. However, given the example, it's about distinct digits.

Alternatively, 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, regardless of position.

But in that case, the answer would be the number of digits that are present in both representations.

But in the example, 121_3 and 413_5 have digits 1,2,1 and 4,1,3. The digits present in both are 1. So answer is 1.

Therefore, the answer is the number of distinct digits that are present in both representations.

Therefore, in our case, since both representations have digits 1,5,6, the answer is 3.

But let me check if there is any other digit. For example, in base 7, digits are 1,5,5,6. In base 8, digits are 1,1,6,5. So digits are 1,5,6.

Therefore, answer is 3.

But maybe the problem is considering digits that are same in both representations, but considering the digits in the numbers. For example, if a digit is present in both numbers, even if it's not in the same position.

But that's what we did.

Alternatively, maybe the problem is asking for the number of digits that are same in both representations, considering the digits in the same position.

But in that case, for our problem:

Base 7: 1 5 5 6

Base 8: 1 1 6 5

Comparing each 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, there are no digits in the same position. Therefore, the example contradicts that interpretation.

Therefore, the problem must be referring to digits that are present in both representations, regardless of their positions.

Therefore, the answer is 3.

But let me check once more. Maybe the problem is from a source where "same digits" refers to digits that are same in value, not symbols. For example, if a digit in base 7 is equal to a digit in base 8. But since digits are symbols, they are same if they have the same symbol.

But if we consider the numeric value, then digits in base 7 can be 0-6, and digits in base 8 can be 0-7. However, the digits present in the numbers are 1,5,6 in both. So their numeric values are same.

But again, the problem says "same digits", which would refer to the symbols.

Therefore, the answer is 3.

But maybe I need to check if there is a mistake in the problem's example. Let me check the example again.

The example says: 121_3 and 413_5 have one digit in common.

In base 3, 121 has digits 1, 2, 1. In base 5, 413 has digits 4, 1, 3. The common digit is 1. So only one digit.

Therefore, the answer is the number of distinct digits that are common.

Therefore, in our problem, since there are three common digits, the answer is 3.

But maybe the problem is from a source where digits are considered as numbers, not symbols. For example, if a digit in base 7 is 1, which is same as digit 1 in base 8. So same numeric value.

But in that case, the digits are same.

But in that case, the answer is still 3.

Therefore, I think the answer is 3.

But to be thorough, let me check if there is any other digit that is common.

In base 7 digits: 1,5,5,6.

In base 8 digits: 1,1,6,5.

So digits are 1,5,6.

No other digits.

Therefore, answer is 3.

But maybe the problem is in the way digits are counted. For example, if digits are considered with multiplicity.

For example, in base 7, there are two 5s and one 1 and one 6. In base 8, there are two 1s, one 6, and one 5.

So the digits that are same in both representations, considering multiplicity, would be:

For digit 1: appears once in base 7 and twice in base 8. So total same digits? If we count the minimum, it's 1.

For digit 5: appears twice in base 7 and once in base 8. So minimum is 1.

For digit 6: appears once in base 7 and once in base 8. So minimum is 1.

Total same digits: 1+1+1=3.

But if we count the maximum, it would be 2+1+1=4. But the example says that in the case of 121_3 and 413_5, which have digits 1,2,1 and 4,1,3, the common digits are 1. So if we count the minimum, it's 1.

Therefore, the answer is 3.

Therefore, I think the answer is 3.

But let me check once more.

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 numbers, regardless of how many times.

But that is the same as the number of distinct digits.

Therefore, answer is 3.

But maybe the problem is from a source where digits are considered as numbers, and the question is about how many digits (as numbers) are same in both representations.

But in that case, the digits are 1,5,6 in both, so three digits.

Therefore, answer is 3.

But maybe I need to check if there is a mistake in my conversion.

Wait, let me check base 7 again.

If I convert 1556 base 7 to decimal:

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

Base 8: 1165 base 8 is 1*8^3 + 1*8^2 + 6*8 + 5 = 512 + 64 + 48 + 5 = 512 + 64 is 576, 576 + 48 is 624, 624 + 5 is 629. Correct.

Therefore, the digits are correct.

Therefore, the answer is 3.

But maybe the problem is in the way digits are written. For example, in base 7, the digits are 1,5,5,6. In base 8, digits are 1,1,6,5. So if we list all digits in both representations, considering multiplicity, then:

Base 7 digits: [1,5,5,6]

Base 8 digits: [1,1,6,5]

The digits that are common are 1,5,6.

But if we count how many times they appear in both:

For digit 1: appears once in base 7 and twice in base 8.

For digit 5: appears twice in base 7 and once in base 8.

For digit 6: appears once in base 7 and once in base 8.

But the problem says "how many of the same digits are found". If it's asking for the total count of digits that are same in both representations, considering multiplicity, then it would be the sum of the minimum counts for each digit.

For digit 1: min(1,2)=1

For digit 5: min(2,1)=1

For digit 6: min(1,1)=1

Total: 1+1+1=3.

Alternatively, if it's asking for the total number of digits that are same in both representations, considering that each occurrence is counted, then it would be the sum of the minimum counts.

But the example given in the problem says that 121_3 and 413_5 have one digit in common. In that case, the digits in base 3 are [1,2,1], and in base 5 are [4,1,3]. The common digits are 1.

If we count the minimum counts: digit 1 appears twice in base 3 and once in base 5, so min(2,1)=1. Therefore, total is 1. Which matches the example.

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 interpretation.

Another possible interpretation: the problem is asking for the number of digits that are same in both representations, considering the digits in the same position.

For example, if a digit is same in the same position in both numbers, then it's counted.

In our case:

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 case, there are no digits in the same position. Therefore, this interpretation is invalid.

Therefore, the answer must be 3.

Therefore, I think the answer is 3.

But just to be thorough, let me check if there is any other digit that is common.

In base 7 digits: 1,5,5,6.

In base 8 digits: 1,1,6,5.

So digits are 1,5,6.

No other digits.

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 into 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, we get:

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

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, we get:

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

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.

---

### Step 4: Interpret the Question

The question asks:
> "How many of the same digits are found in the base 7 and base 8 representations of 629?"

This is interpreted as asking for the **number of distinct digits** that are **common** to both representations. Since the digits **1**, **5**, and **6** are all present in both, the answer is:

$$
\boxed{3}
$$

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 different 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 of these combinations, I need to assign the 3 people to the 3 selected offices. Since the offices are different, the assignment matters. So, for each combination of offices, the number of assignments is the number of permutations of 3 people in 3 offices, which is 3! (3 factorial).

Calculating that: C(5,3) is 10, and 3! is 6. So, 10 * 6 = 60.

So both approaches give me 60. That seems consistent. Therefore, the answer should be 60.

Wait, let me check if there's another way to think about this. Maybe 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, that's the same answer.

So, all three methods—sequential assignment, combination with permutation, and direct permutation formula—give me 60. That gives me more confidence that the answer is correct.

But let me think again if there's any chance I might have overcounted or undercounted.

Suppose I have offices labeled 1, 2, 3, 4, 5. Let's say the people are Alice, Bob, and Charlie.

If I assign Alice to office 1, Bob to office 2, and Charlie to office 3, that's one arrangement.

Alternatively, if I assign Alice to office 2, Bob to office 1, and Charlie to office 3, that's a different arrangement. Since the offices are different, these are distinct.

So, each different assignment is unique. Therefore, the count should be correct.

Alternatively, if the offices were identical, then it would be a combination problem, but since they are different, permutations are needed.

Another way to think about it: For each person, the number of choices decreases by one each time. So, first person has 5 choices, second has 4, third has 3. Multiply them together: 5*4*3=60.

Alternatively, if the offices were not distinct, but the people are, then it would be combinations. But since offices are distinct, it's permutations.

Wait, actually, the problem says "assign to 5 different offices". So, each office is unique. Therefore, assigning different people to different offices is a permutation.

Alternatively, if the problem had said that the offices are identical, then it would be combinations. But since they are different, permutations.

Therefore, the answer is 5*4*3=60.

Alternatively, if I think of it as functions. Each person is mapped to an office, with the condition that no two people are mapped to the same office. So, the number of injective functions from the set of 3 people to the set of 5 offices.

The number of injective functions is P(5,3)=60.

Yes, that's another way to think about it.

So, all these different perspectives lead me to the same answer. Therefore, I think the answer is 60.

But just to make sure, let me try a smaller case. Let's say there are 2 people and 3 offices. How many ways?

Using the same logic: 3*2=6.

Alternatively, combinations: C(3,2)=3, then 2! =2, so 3*2=6.

Listing them out:

People A and B. Offices 1,2,3.

Possible assignments:

A1, B2

A1, B3

A2, B1

A2, B3

A3, B1

A3, B2

Total of 6. Which matches. So, that seems correct.

Therefore, scaling up, with 3 people and 5 offices, it's 5*4*3=60.

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 two people share the same office, we can approach the problem using the concept of **permutations**.

---

### Understanding the Problem

We are given:

- **3 people** (let's call them A, B, and C)
- **5 different offices** (say, labeled 1 through 5)
- **Each person must be assigned to exactly one office**
- **No two people can share the same office**

This means we are looking for the number of **injective mappings** from the set of 3 people to the set of 5 offices. In other words, we are choosing 3 distinct offices out of 5 and assigning each person to one of them, with no repetition.

---

### Step-by-Step Solution

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

1. **First person** (say, A) can be assigned to any of the 5 offices.
2. **Second person** (B) can be assigned to any of the remaining 4 offices.
3. **Third person** (C) can be assigned to any of the remaining 3 offices.

So, the total number of assignments is:

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

---

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

1. **Choose 3 offices** from the 5 available. The number of ways to do this is:

$$
\binom{5}{3} = 10
$$

2. **Assign the 3 people** to these 3 selected offices. Since the offices are distinct, the number of ways to assign 3 people to 3 offices is:

$$
3! = 6
$$

3. Multiply the two results:

$$
10 \times 6 = 60
$$

---

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

This is a direct application of the permutation formula:

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

---

### Final Answer

All three methods lead to the same result. Therefore, the number of ways to assign 3 people to 5 different offices, with no two people sharing the same 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 different 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 of these combinations, I need to assign the 3 people to the 3 selected offices. Since the offices are different, the assignment matters. So, for each combination of offices, the number of assignments is the number of permutations of 3 people in 3 offices, which is 3! (3 factorial).

Calculating that: C(5,3) is 10, and 3! is 6. So, 10 * 6 = 60.

So both approaches give me 60. That seems consistent. Therefore, the answer should be 60.

Wait, let me check if there's another way to think about this. Maybe 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, that's the same answer.

So, all three methods—sequential assignment, combination with permutation, and direct permutation formula—give me 60. That gives me more confidence that the answer is correct.

But let me think again if there's any chance I might have overcounted or undercounted.

Suppose I have offices labeled 1, 2, 3, 4, 5. Let's say the people are Alice, Bob, and Charlie.

If I assign Alice to office 1, Bob to office 2, and Charlie to office 3, that's one arrangement.

Alternatively, if I assign Alice to office 2, Bob to office 1, and Charlie to office 3, that's a different arrangement. Since the offices are different, these are distinct.

So, each different assignment is unique. Therefore, the count should be correct.

Alternatively, if the offices were identical, then it would be a combination problem, but since they are different, permutations are needed.

Another way to think about it: For each person, the number of choices decreases by one each time. So, first person has 5 choices, second has 4, third has 3. Multiply them together: 5*4*3=60.

Alternatively, if the offices were not distinct, but the people are, then it would be combinations. But since offices are distinct, it's permutations.

Wait, actually, the problem says "assign to 5 different offices". So, each office is unique. Therefore, assigning different people to different offices is a permutation.

Alternatively, if the problem had said that the offices are identical, then it would be combinations. But since they are different, permutations.

Therefore, the answer is 5*4*3=60.

Alternatively, if I think of it as functions. Each person is mapped to an office, with the condition that no two people are mapped to the same office. So, the number of injective functions from the set of 3 people to the set of 5 offices.

The number of injective functions is P(5,3)=60.

Yes, that's another way to think about it.

So, all these different perspectives lead me to the same answer. Therefore, I think the answer is 60.

But just to make sure, let me try a smaller case. Let's say there are 2 people and 3 offices. How many ways?

Using the same logic: 3*2=6.

Alternatively, combinations: C(3,2)=3, then 2! =2, so 3*2=6.

Listing them out:

People A and B. Offices 1,2,3.

Possible assignments:

A1, B2

A1, B3

A2, B1

A2, B3

A3, B1

A3, B2

Total of 6. Which matches. So, that seems correct.

Therefore, scaling up, with 3 people and 5 offices, it's 5*4*3=60.

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 two people share the same office, we can approach the problem using the concept of **permutations**.

---

### Understanding the Problem

We are given:

- **3 people** (let's call them A, B, and C)
- **5 different offices** (say, labeled 1 through 5)
- **Each person must be assigned to exactly one office**
- **No two people can share the same office**

This means we are looking for the number of **injective mappings** from the set of 3 people to the set of 5 offices. In other words, we are choosing 3 distinct offices out of 5 and assigning each person to one of them, with no repetition.

---

### Step-by-Step Solution

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

1. **First person** (say, A) can be assigned to any of the 5 offices.
2. **Second person** (B) can be assigned to any of the remaining 4 offices.
3. **Third person** (C) can be assigned to any of the remaining 3 offices.

So, the total number of assignments is:

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

---

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

1. **Choose 3 offices** from the 5 available. The number of ways to do this is:

$$
\binom{5}{3} = 10
$$

2. **Assign the 3 people** to these 3 selected offices. Since the offices are distinct, the number of ways to assign 3 people to 3 offices is:

$$
3! = 6
$$

3. Multiply the two results:

$$
10 \times 6 = 60
$$

---

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

This is a direct application of the permutation formula:

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

---

### Final Answer

All three methods lead to the same result. Therefore, the number of ways to assign 3 people to 5 different offices, with no two people sharing the same 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, but that seems different from what I did before. Wait, hold on. Let me check that again.

Wait, if I have x = 2/(y + 1), then solving for y:

Multiply both sides by (y + 1):

x(y + 1) = 2

Which is x*y + x = 2

Then, subtract x:

x*y = 2 - x

Then, divide both sides by x:

y = (2 - x)/x

But that can also 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)/x or 2/x - 1.

Therefore, if I want to find f⁻¹(1/5), I can plug 1/5 into this inverse function.

So let's compute that:

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 that's consistent with my previous result.

Alternatively, using the other form of the inverse function: (2 - x)/x. So plugging in x = 1/5:

(2 - 1/5)/(1/5) = (10/5 - 1/5)/(1/5) = (9/5)/(1/5) = 9/5 * 5/1 = 9.

Same result. So that's good.

Therefore, both methods give me 9, and plugging back into the original function confirms that f(9) = 1/5.

Wait, but let me just make sure that there isn't a mistake in the inverse function derivation. Let me check with another value. For example, suppose I take 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. Wait, but x = 0 is not in the domain of f(x) because f(0) = 2/(0 + 1) = 2. So f(0) = 2. Then f⁻¹(2) should be 0. Let me check:

f⁻¹(2) = 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 = 2*(3/2) - 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 inverse function graphically. The original function f(x) = 2/(x + 1) is a hyperbola, and its inverse should be the reflection over the line y = x. So if I have a point (a, b) on f(x), then the inverse function will have the point (b, a).

So, for example, if I have the point (9, 1/5) on f(x), then the inverse function will have the point (1/5, 9). Therefore, f⁻¹(1/5) is 9.

But again, this is just another way of thinking about it, but it still leads to the same conclusion.

Is there any chance that I could have messed up the algebra? Let me go through the inverse function steps again.

Original function: y = 2/(x + 1).

Swap x and y: x = 2/(y + 1).

Solve for y: Multiply both sides by (y + 1): x(y + 1) = 2.

Expand: xy + x = 2.

Subtract x: xy = 2 - x.

Divide by x: y = (2 - x)/x.

Yes, that's correct.

Alternatively, maybe I can write it as y = 2/x - 1.

Yes, because (2 - x)/x = 2/x - x/x = 2/x - 1.

So, that's correct.

Therefore, plugging in 1/5 into this inverse function:

y = 2/(1/5) - 1 = 10 - 1 = 9.

So, all steps check out.

Alternatively, if I wanted to write the inverse function in terms of f⁻¹(x), then f⁻¹(x) = (2 - x)/x.

But maybe I can also write it as f⁻¹(x) = 2/x - 1.

Either way, plugging in x = 1/5 gives 9.

Therefore, I think I can be confident that the answer is 9.

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).

Which is 10 = x + 1.

Subtract 1: x = 9.

Yep, same result.

So, all methods lead to x = 9.

Therefore, the value of f⁻¹(1/5) is 9.

I think that's solid. I don't see any mistakes in my reasoning.

**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 other words, we are looking for the input $ x $ that maps to $ \frac{1}{5} $ under the function $ f(x) = \frac{2}{x + 1} $.

---

### Step 1: Set up the equation

We are given:
$$
f(x) = \frac{2}{x + 1}
$$
We want to find $ x $ such that:
$$
f(x) = \frac{1}{5}
$$
So we set:
$$
\frac{2}{x + 1} = \frac{1}{5}
$$

---

### Step 2: Solve for $ x $

To solve this equation, we can cross-multiply:
$$
2 \cdot 5 = 1 \cdot (x + 1)
$$
$$
10 = x + 1
$$
$$
x = 10 - 1 = 9
$$

---

### Step 3: Verify the result

Let’s verify that $ f(9) = \frac{1}{5} $:
$$
f(9) = \frac{2}{9 + 1} = \frac{2}{10} = \frac{1}{5}
$$
This confirms that $ f(9) = \frac{1}{5} $, so $ f^{-1}\left(\frac{1}{5}\right) = 9 $.

---

### Step 4: Alternative method using inverse function

To further confirm, we can derive the inverse function explicitly.

Start with:
$$
y = \frac{2}{x + 1}
$$
Swap $ x $ and $ y $:
$$
x = \frac{2}{y + 1}
$$
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 plug in $ x = \frac{1}{5} $:
$$
f^{-1}\left(\frac{1}{5}\right) = \frac{2 - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{10}{5} - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{9}{5}}{\frac{1}{5}} = 9
$$

---

### 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, but that seems different from what I did before. Wait, hold on. Let me check that again.

Wait, if I have x = 2/(y + 1), then solving for y:

Multiply both sides by (y + 1):

x(y + 1) = 2

Which is x*y + x = 2

Then, subtract x:

x*y = 2 - x

Then, divide both sides by x:

y = (2 - x)/x

But that can also 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)/x or 2/x - 1.

Therefore, if I want to find f⁻¹(1/5), I can plug 1/5 into this inverse function.

So let's compute that:

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 that's consistent with my previous result.

Alternatively, using the other form of the inverse function: (2 - x)/x. So plugging in x = 1/5:

(2 - 1/5)/(1/5) = (10/5 - 1/5)/(1/5) = (9/5)/(1/5) = 9/5 * 5/1 = 9.

Same result. So that's good.

Therefore, both methods give me 9, and plugging back into the original function confirms that f(9) = 1/5.

Wait, but let me just make sure that there isn't a mistake in the inverse function derivation. Let me check with another value. For example, suppose I take 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. Wait, but x = 0 is not in the domain of f(x) because f(0) = 2/(0 + 1) = 2. So f(0) = 2. Then f⁻¹(2) should be 0. Let me check:

f⁻¹(2) = 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 = 2*(3/2) - 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 inverse function graphically. The original function f(x) = 2/(x + 1) is a hyperbola, and its inverse should be the reflection over the line y = x. So if I have a point (a, b) on f(x), then the inverse function will have the point (b, a).

So, for example, if I have the point (9, 1/5) on f(x), then the inverse function will have the point (1/5, 9). Therefore, f⁻¹(1/5) is 9.

But again, this is just another way of thinking about it, but it still leads to the same conclusion.

Is there any chance that I could have messed up the algebra? Let me go through the inverse function steps again.

Original function: y = 2/(x + 1).

Swap x and y: x = 2/(y + 1).

Solve for y: Multiply both sides by (y + 1): x(y + 1) = 2.

Expand: xy + x = 2.

Subtract x: xy = 2 - x.

Divide by x: y = (2 - x)/x.

Yes, that's correct.

Alternatively, maybe I can write it as y = 2/x - 1.

Yes, because (2 - x)/x = 2/x - x/x = 2/x - 1.

So, that's correct.

Therefore, plugging in 1/5 into this inverse function:

y = 2/(1/5) - 1 = 10 - 1 = 9.

So, all steps check out.

Alternatively, if I wanted to write the inverse function in terms of f⁻¹(x), then f⁻¹(x) = (2 - x)/x.

But maybe I can also write it as f⁻¹(x) = 2/x - 1.

Either way, plugging in x = 1/5 gives 9.

Therefore, I think I can be confident that the answer is 9.

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).

Which is 10 = x + 1.

Subtract 1: x = 9.

Yep, same result.

So, all methods lead to x = 9.

Therefore, the value of f⁻¹(1/5) is 9.

I think that's solid. I don't see any mistakes in my reasoning.

**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 other words, we are looking for the input $ x $ that maps to $ \frac{1}{5} $ under the function $ f(x) = \frac{2}{x + 1} $.

---

### Step 1: Set up the equation

We are given:
$$
f(x) = \frac{2}{x + 1}
$$
We want to find $ x $ such that:
$$
f(x) = \frac{1}{5}
$$
So we set:
$$
\frac{2}{x + 1} = \frac{1}{5}
$$

---

### Step 2: Solve for $ x $

To solve this equation, we can cross-multiply:
$$
2 \cdot 5 = 1 \cdot (x + 1)
$$
$$
10 = x + 1
$$
$$
x = 10 - 1 = 9
$$

---

### Step 3: Verify the result

Let’s verify that $ f(9) = \frac{1}{5} $:
$$
f(9) = \frac{2}{9 + 1} = \frac{2}{10} = \frac{1}{5}
$$
This confirms that $ f(9) = \frac{1}{5} $, so $ f^{-1}\left(\frac{1}{5}\right) = 9 $.

---

### Step 4: Alternative method using inverse function

To further confirm, we can derive the inverse function explicitly.

Start with:
$$
y = \frac{2}{x + 1}
$$
Swap $ x $ and $ y $:
$$
x = \frac{2}{y + 1}
$$
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 plug in $ x = \frac{1}{5} $:
$$
f^{-1}\left(\frac{1}{5}\right) = \frac{2 - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{10}{5} - \frac{1}{5}}{\frac{1}{5}} = \frac{\frac{9}{5}}{\frac{1}{5}} = 9
$$

---

### 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 $x=\boxed{9}$.