Explanation:
Okay, so I need to convert the binary number 10101001110₂ to octal. Hmm, let me remember how to do this. I think there's a method where you can group the binary digits into sets of three, starting from the right, and then convert each group to its octal equivalent. Let me try to recall the exact steps.

First, binary numbers are base 2, and octal is base 8. Since 8 is 2³, each octal digit corresponds to three binary digits. That makes sense. So, if I can split the binary number into groups of three bits, each group can be converted to an octal digit. But I need to make sure that the number of digits is a multiple of three. If not, I might have to add leading zeros to the leftmost group.

Let me check the given binary number: 10101001110. Let me count the digits. Let's see: 1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. Hmm, 11 divided by 3 is 3 with a remainder of 2. So, there are two digits left over on the left side. To make it a multiple of three, I need to add one leading zero. Wait, adding one leading zero would make it 12 digits? Wait, 11 digits plus one zero would be 12 digits, which is divisible by three. Let me confirm:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Adding a leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Now, grouping into threes from the right. Wait, actually, when grouping for octal, you start from the right, right? So maybe I should split it into groups of three starting from the right. Let me try that.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate the groups. Starting from the right:

First, let me count the digits again. Let me write them down:

Position: 1 2 3 4 5 6 7 8 9 10 11

Digits:    1 0 1 0 1 0 0 1 1 1 0

Starting from the right (position 11), group into threes:

Group 1 (rightmost): positions 9-11: digits 1, 1, 0

Wait, positions 9, 10, 11: digits 1, 1, 0? Wait, position 11 is the rightmost digit. Let me index them properly.

Let me write the binary number as:

Digits: 1 0 1 0 1 0 0 1 1 1 0

Let me number them from right to left as positions 0 to 10:

Wait, maybe it's easier to reverse the number and group from the left. Let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Reversed: 0 1 1 1 0 0 1 0 1 0 1

Now, grouping into threes from the left (since we reversed it):

Group 1: 0 1 1

Group 2: 1 0 0

Group 3: 1 0 1

Wait, but reversed, so original groups would be:

Wait, maybe I confused the direction. Let me think again.

When converting binary to octal, you start grouping from the right. If the number of digits isn't a multiple of three, add leading zeros to the left. So, original number is 10101001110. Let me count the digits again: 1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. So, adding one leading zero to make it 12 digits: 0 1 0 1 0 1 0 0 1 1 1 0.

Now, grouping into threes from the left:

Group 1: 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4: 1 1 0

Wait, let me check:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three:

First group (leftmost): 0 1 0

Second group: 1 0 1

Third group: 0 0 1

Fourth group: 1 1 0

Wait, that's four groups. Let me check the number of digits: 3*4=12, which matches. So each group is three bits. Now, convert each group to octal.

Let me write them down:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Now, converting each binary group to octal:

Group 1: 010. Let me recall that 0*2² + 1*2¹ + 0*2⁰ = 0 + 2 + 0 = 2. So that's 2 in octal.

Group 2: 101. 1*4 + 0*2 + 1*1 = 4 + 0 + 1 = 5. So 5.

Group 3: 001. 0*4 + 0*2 + 1*1 = 1. So 1.

Group 4: 110. 1*4 + 1*2 + 0*1 = 4 + 2 + 0 = 6. So 6.

So putting them together, the octal number is 6 1 5 2? Wait, wait, the groups were from left to right as group 1 to group 4. Wait, group 1 was the leftmost group, which is 010, which is 2, then group 2 is 101 which is 5, group 3 is 001 which is 1, group 4 is 110 which is 6. So combining them, the octal number is 2 5 1 6? Wait, but group 1 is the leftmost group, so the order is group 1, group 2, group 3, group 4. So 2, 5, 1, 6? Wait, but that would be 2516₈? Wait, but let me check again.

Wait, original grouping after adding leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups:

First three bits: 010 (group 1)

Next three: 101 (group 2)

Next three: 001 (group 3)

Next three: 110 (group 4)

So the octal digits are group1 group2 group3 group4: 2 5 1 6. So 2516 in octal? Wait, but let me check if I did the grouping correctly.

Wait, the original number with leading zero is 0 1 0 1 0 1 0 0 1 1 1 0. Let me write them as:

Group 1: 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4: 1 1 0

Yes, that's correct. So converting each group:

Group 1: 010 = 2

Group 2: 101 = 5

Group 3: 001 = 1

Group 4: 110 = 6

So combining them, the octal number is 2 5 1 6, which is 2516₈. Wait, but let me check if that's correct. Alternatively, maybe I made a mistake in grouping.

Alternatively, maybe I should have grouped from the right. Let me try that method again.

Original binary number: 10101001110. Let me count the digits again: 1 0 1 0 1 0 0 1 1 1 0. 11 digits. So, to group into threes from the right, we need to add leading zeros to make the total number of digits a multiple of three. Since 11 divided by 3 is 3 with remainder 2, we need to add one leading zero. So adding one zero to the left, making it 12 digits: 0 1 0 1 0 1 0 0 1 1 1 0.

Now, grouping into threes from the right:

Starting from the right, the groups would be:

Group 1: last three digits: 1 1 0

Group 2: previous three: 0 0 1

Group 3: previous three: 1 0 1

Group 4: remaining three: 0 1 0

Wait, so grouping from the right:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups from the right:

Group 1 (rightmost): positions 9-11: digits 1, 1, 0

Group 2: positions 6-8: digits 0, 0, 1

Group 3: positions 3-5: digits 1, 0, 1

Group 4: positions 0-2: digits 0, 1, 0

Wait, but positions are confusing. Let me write the number with spaces:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups of three starting from the right:

Starting from the right, the last three digits are 1 1 0 (positions 9-11 if we count from 0). Then the next three digits to the left are 0 0 1 (positions 6-8). Then the next three digits are 1 0 1 (positions 3-5). Then the remaining digits are 0 1 0 (positions 0-2). So the groups are:

Group 1: 110

Group 2: 001

Group 3: 101

Group 4: 010

So converting each group:

Group 1: 110 = 6

Group 2: 001 = 1

Group 3: 101 = 5

Group 4: 010 = 2

So combining them from left to right (since group 4 is the leftmost group), the octal number is 2 5 1 6, which is 2516₈. So same result as before. So that seems consistent.

But let me verify this by converting the binary number to decimal first, then to octal, to check if it's the same. Maybe that's a good idea to confirm.

First, converting 10101001110₂ to decimal.

Let me write down the binary number: 1 0 1 0 1 0 0 1 1 1 0.

Let me write the positions from right to left, starting at 0:

Position: 10 9 8 7 6 5 4 3 2 1 0

Wait, but the number is 11 digits, so positions 10 down to 0.

Wait, let me write the binary number as:

From left to right: positions 10 to 0:

Digit positions:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, let me check:

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

So leftmost digit is position 10 (since there are 11 digits), and rightmost is position 0.

So the value is:

1*(2^10) + 0*(2^9) + 1*(2^8) + 0*(2^7) + 1*(2^6) + 0*(2^5) + 0*(2^4) + 1*(2^3) + 1*(2^2) + 1*(2^1) + 0*(2^0)

Calculating each term:

2^10 = 1024

2^8 = 256

2^6 = 64

2^3 = 8

2^2 = 4

2^1 = 2

So adding them up:

1*1024 = 1024

0*512 = 0

1*256 = 256

0*128 = 0

1*64 = 64

0*32 = 0

0*16 = 0

1*8 = 8

1*4 = 4

1*2 = 2

0*1 = 0

Adding all these together:

1024 + 256 = 1280

1280 + 64 = 1344

1344 + 8 = 1352

1352 + 4 = 1356

1356 + 2 = 1358

So decimal value is 1358.

Now, converting 1358 decimal to octal.

To convert decimal to octal, divide by 8 and find remainders.

Let me do that:

1358 ÷ 8 = 169.75, so quotient 169, remainder 1358 - 169*8 = 1358 - 1352 = 6.

So first remainder (rightmost digit): 6.

169 ÷ 8 = 21.125, quotient 21, remainder 169 - 21*8 = 169 - 168 = 1.

Second remainder: 1.

21 ÷ 8 = 2.625, quotient 2, remainder 21 - 16 = 5.

Third remainder: 5.

2 ÷ 8 = 0.25, quotient 0, remainder 2.

Fourth remainder: 2.

So writing the remainders from last to first: 2, 5, 1, 6. So octal number is 2516₈. Which matches the previous result. So that's good. So the answer is 2516 in octal.

But let me check once more with another method. Maybe converting binary to octal directly by another grouping.

Alternatively, since each octal digit is three bits, maybe I can split the original binary number into groups of three, but without adding leading zeros. Wait, but the original number has 11 bits. If I group from the right, as we did before, adding one leading zero, then we get 2516. But maybe I can also split the original number into groups of three, starting from the right, and pad with leading zeros if necessary.

Original binary number: 10101001110

Split into groups of three from the right:

Starting from the right, the last three digits are 110, then the next three digits are 001, then the next three digits are 101, and then the remaining digits are 010. Wait, that's the same as before. So 010 101 001 110, which is 2 5 1 6. So same result.

Alternatively, if I didn't add the leading zero, but tried to group from the left, but that would be problematic because the leftmost group would have only two bits. So adding leading zeros is necessary.

Alternatively, maybe I can convert the binary number to hexadecimal first, then to octal? Let me try that as another check.

First, converting binary to hexadecimal. Since hexadecimal is base 16, which is 2⁴, so group into fours.

Original binary number: 10101001110. Let me count the digits again: 11 digits. To group into fours, need to add leading zeros to make it a multiple of four. 11 divided by 4 is 2 with remainder 3, so add one leading zero to make it 12 digits. Wait, 11 + 1 = 12, which is divisible by 4? No, 12 is divisible by 4? 12 / 4 = 3. Yes. So adding one leading zero: 0 1 0 1 0 1 0 0 1 1 1 0.

Now, grouping into fours from the left:

Group 1: 0101

Group 2: 0100

Group 3: 1110

Wait, let me check:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of four:

Group 1: 0101

Group 2: 0100

Group 3: 1110

Wait, but that's three groups. Let me check the digits:

Group 1: first four digits: 0 1 0 1

Group 2: next four: 0 1 0 0

Group 3: next four: 1 1 1 0

Yes. Now converting each group to hexadecimal:

Group 1: 0101 is 5 in hex (since 0*8 + 1*4 + 0*2 + 1*1 = 5)

Group 2: 0100 is 4 in hex (0*8 + 1*4 + 0*2 + 0*1 = 4)

Group 3: 1110 is E in hex (1*8 + 1*4 + 1*2 + 0*1 = 14, which is E)

So hexadecimal number is 5 4 E, which is 54E₁₆.

Now, converting 54E₁₆ to octal. How to do that? Well, one way is to convert hexadecimal to decimal first, then decimal to octal. But we already know that 54E₁₆ is equal to 1358 decimal, which we found earlier. So converting 1358 decimal to octal gives 2516, which matches. Alternatively, maybe convert hexadecimal to octal directly?

But since hexadecimal is base 16 and octal is base 8, which are not powers of the same base, it's not straightforward. However, since 16 is 2⁴ and 8 is 2³, maybe we can convert each hexadecimal digit to 4 binary digits, then group into 3s for octal. Let me try that.

So 54E₁₆ is 0101 0100 1110 in binary. Wait, that's the same as before. Then, converting that to octal by grouping into 3s. Let me take the binary number 010101001110. Let me add leading zeros if necessary. Wait, the binary number is 12 bits: 010101001110. Let me group into 3s from the right:

Original binary: 010101001110

Split into groups of three from the right:

Starting from the right:

Group 1: 110

Group 2: 001

Group 3: 101

Group 4: 010

Which is the same as before. So converting each group:

110 = 6, 001 = 1, 101 = 5, 010 = 2. So octal number is 2516. Same result. So that's consistent.

Therefore, all methods lead to the octal number 2516. Therefore, I can be confident that the answer is 2516 in octal.

But just to make sure, let me check once again by converting the binary number to decimal and then to octal, as I did before.

We had the binary number 10101001110₂ equals 1358 decimal. Converting 1358 to octal:

Divide 1358 by 8:

1358 ÷ 8 = 169.75 → quotient 169, remainder 6

169 ÷ 8 = 21.125 → quotient 21, remainder 1

21 ÷ 8 = 2.625 → quotient 2, remainder 5

2 ÷ 8 = 0.25 → quotient 0, remainder 2

Reading the remainders from last to first: 2, 5, 1, 6 → 2516₈. Yep, same result.

Therefore, I think the answer is definitely 2516 in octal.

But just to make sure I didn't make any mistake in grouping, let me write the binary number again with the added leading zero and group them:

Original binary: 10101001110

Add leading zero: 010101001110

Split into groups of three from the right:

Starting from the right:

First group: 110 (positions 0-2)

Second group: 001 (positions 3-5)

Third group: 101 (positions 6-8)

Fourth group: 010 (positions 9-11)

Wait, positions are confusing, but if I write the binary number as:

0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three from the right:

Group 1: last three digits: 1 1 0 → 110

Group 2: previous three: 0 0 1 → 001

Group 3: previous three: 1 0 1 → 101

Group 4: remaining three: 0 1 0 → 010

So converting each group:

Group 1: 110 = 6

Group 2: 001 = 1

Group 3: 101 = 5

Group 4: 010 = 2

So octal digits are 2 (from group 4), 5 (group 3), 1 (group 2), 6 (group 1). So 2 5 1 6 → 2516. Correct.

Alternatively, if I write the groups in order from left to right, which is group 4, group 3, group 2, group 1. Wait, no. When grouping from the right, the first group is the rightmost, so when writing the octal number, the leftmost group is the leftmost group of the binary number. Wait, maybe I need to think of the groups as follows:

Original binary with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Split into groups of three:

Group 1 (leftmost): 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4 (rightmost): 1 1 0

So converting each group:

Group 1: 010 = 2

Group 2: 101 = 5

Group 3: 001 = 1

Group 4: 110 = 6

So octal number is 2 5 1 6 → 2516. Same result. So regardless of the grouping direction, as long as I'm consistent, I get the same answer.

Therefore, I can be confident that the octal equivalent of the binary number 10101001110₂ is 2516₈.

**Final Answer**
The binary number $10101001110_{2}$ is equal to \boxed{2516} in base eight.

To convert the binary number $10101001110_2$ to its octal equivalent, we can use the fact that **each octal digit corresponds to exactly three binary digits**. This is because $8 = 2^3$, so we can group the binary digits into sets of three, starting from the right, and then convert each group to its octal equivalent.

---

### Step 1: Count the number of binary digits

The binary number is:

$$
10101001110
$$

This has **11 digits**. Since 11 is not a multiple of 3, we need to **add one leading zero** to make the total number of digits a multiple of 3:

$$
010101001110
$$

Now we have **12 digits**, which is divisible by 3.

---

### Step 2: Group the binary digits into sets of three

Starting from the **right**, we group the binary digits into sets of three:

$$
010\ 101\ 001\ 110
$$

---

### Step 3: Convert each group to octal

Now, convert each 3-bit binary group to its octal equivalent:

- $010_2 = 2_8$
- $101_2 = 5_8$
- $001_2 = 1_8$
- $110_2 = 6_8$

---

### Step 4: Combine the octal digits

Putting the octal digits together in the same order as the groups:

$$
2\ 5\ 1\ 6
$$

So, the octal representation is:

$$
\boxed{2516}
$$

---

### Final Answer

$$
\boxed{2516}
$$

[Assistant]
Explanation:
Okay, so I need to convert the binary number 10101001110₂ to octal. Hmm, let me remember how to do this. I think there's a method where you can group the binary digits into sets of three, starting from the right, and then convert each group to its octal equivalent. Let me try to recall the exact steps.

First, binary numbers are base 2, and octal is base 8. Since 8 is 2³, each octal digit corresponds to three binary digits. That makes sense. So, if I can split the binary number into groups of three bits, each group can be converted to an octal digit. But I need to make sure that the number of digits is a multiple of three. If not, I might have to add leading zeros to the leftmost group.

Let me check the given binary number: 10101001110. Let me count the digits. Let's see: 1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. Hmm, 11 divided by 3 is 3 with a remainder of 2. So, there are two digits left over on the left side. To make it a multiple of three, I need to add one leading zero. Wait, adding one leading zero would make it 12 digits? Wait, 11 digits plus one zero would be 12 digits, which is divisible by three. Let me confirm:

Original number: 1 0 1 0 1 0 0 1 1 1 0

Adding a leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Now, grouping into threes from the right. Wait, actually, when grouping for octal, you start from the right, right? So maybe I should split it into groups of three starting from the right. Let me try that.

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

Let me write it with spaces to separate the groups. Starting from the right:

First, let me count the digits again. Let me write them down:

Position: 1 2 3 4 5 6 7 8 9 10 11

Digits:    1 0 1 0 1 0 0 1 1 1 0

Starting from the right (position 11), group into threes:

Group 1 (rightmost): positions 9-11: digits 1, 1, 0

Wait, positions 9, 10, 11: digits 1, 1, 0? Wait, position 11 is the rightmost digit. Let me index them properly.

Let me write the binary number as:

Digits: 1 0 1 0 1 0 0 1 1 1 0

Let me number them from right to left as positions 0 to 10:

Wait, maybe it's easier to reverse the number and group from the left. Let me try that.

Original number: 1 0 1 0 1 0 0 1 1 1 0

Reversed: 0 1 1 1 0 0 1 0 1 0 1

Now, grouping into threes from the left (since we reversed it):

Group 1: 0 1 1

Group 2: 1 0 0

Group 3: 1 0 1

Wait, but reversed, so original groups would be:

Wait, maybe I confused the direction. Let me think again.

When converting binary to octal, you start grouping from the right. If the number of digits isn't a multiple of three, add leading zeros to the left. So, original number is 10101001110. Let me count the digits again: 1 0 1 0 1 0 0 1 1 1 0. That's 11 digits. So, adding one leading zero to make it 12 digits: 0 1 0 1 0 1 0 0 1 1 1 0.

Now, grouping into threes from the left:

Group 1: 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4: 1 1 0

Wait, let me check:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three:

First group (leftmost): 0 1 0

Second group: 1 0 1

Third group: 0 0 1

Fourth group: 1 1 0

Wait, that's four groups. Let me check the number of digits: 3*4=12, which matches. So each group is three bits. Now, convert each group to octal.

Let me write them down:

Group 1: 010

Group 2: 101

Group 3: 001

Group 4: 110

Now, converting each binary group to octal:

Group 1: 010. Let me recall that 0*2² + 1*2¹ + 0*2⁰ = 0 + 2 + 0 = 2. So that's 2 in octal.

Group 2: 101. 1*4 + 0*2 + 1*1 = 4 + 0 + 1 = 5. So 5.

Group 3: 001. 0*4 + 0*2 + 1*1 = 1. So 1.

Group 4: 110. 1*4 + 1*2 + 0*1 = 4 + 2 + 0 = 6. So 6.

So putting them together, the octal number is 6 1 5 2? Wait, wait, the groups were from left to right as group 1 to group 4. Wait, group 1 was the leftmost group, which is 010, which is 2, then group 2 is 101 which is 5, group 3 is 001 which is 1, group 4 is 110 which is 6. So combining them, the octal number is 2 5 1 6? Wait, but group 1 is the leftmost group, so the order is group 1, group 2, group 3, group 4. So 2, 5, 1, 6? Wait, but that would be 2516₈? Wait, but let me check again.

Wait, original grouping after adding leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups:

First three bits: 010 (group 1)

Next three: 101 (group 2)

Next three: 001 (group 3)

Next three: 110 (group 4)

So the octal digits are group1 group2 group3 group4: 2 5 1 6. So 2516 in octal? Wait, but let me check if I did the grouping correctly.

Wait, the original number with leading zero is 0 1 0 1 0 1 0 0 1 1 1 0. Let me write them as:

Group 1: 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4: 1 1 0

Yes, that's correct. So converting each group:

Group 1: 010 = 2

Group 2: 101 = 5

Group 3: 001 = 1

Group 4: 110 = 6

So combining them, the octal number is 2 5 1 6, which is 2516₈. Wait, but let me check if that's correct. Alternatively, maybe I made a mistake in grouping.

Alternatively, maybe I should have grouped from the right. Let me try that method again.

Original binary number: 10101001110. Let me count the digits again: 1 0 1 0 1 0 0 1 1 1 0. 11 digits. So, to group into threes from the right, we need to add leading zeros to make the total number of digits a multiple of three. Since 11 divided by 3 is 3 with remainder 2, we need to add one leading zero. So adding one zero to the left, making it 12 digits: 0 1 0 1 0 1 0 0 1 1 1 0.

Now, grouping into threes from the right:

Starting from the right, the groups would be:

Group 1: last three digits: 1 1 0

Group 2: previous three: 0 0 1

Group 3: previous three: 1 0 1

Group 4: remaining three: 0 1 0

Wait, so grouping from the right:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups from the right:

Group 1 (rightmost): positions 9-11: digits 1, 1, 0

Group 2: positions 6-8: digits 0, 0, 1

Group 3: positions 3-5: digits 1, 0, 1

Group 4: positions 0-2: digits 0, 1, 0

Wait, but positions are confusing. Let me write the number with spaces:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Let me separate into groups of three starting from the right:

Starting from the right, the last three digits are 1 1 0 (positions 9-11 if we count from 0). Then the next three digits to the left are 0 0 1 (positions 6-8). Then the next three digits are 1 0 1 (positions 3-5). Then the remaining digits are 0 1 0 (positions 0-2). So the groups are:

Group 1: 110

Group 2: 001

Group 3: 101

Group 4: 010

So converting each group:

Group 1: 110 = 6

Group 2: 001 = 1

Group 3: 101 = 5

Group 4: 010 = 2

So combining them from left to right (since group 4 is the leftmost group), the octal number is 2 5 1 6, which is 2516₈. So same result as before. So that seems consistent.

But let me verify this by converting the binary number to decimal first, then to octal, to check if it's the same. Maybe that's a good idea to confirm.

First, converting 10101001110₂ to decimal.

Let me write down the binary number: 1 0 1 0 1 0 0 1 1 1 0.

Let me write the positions from right to left, starting at 0:

Position: 10 9 8 7 6 5 4 3 2 1 0

Wait, but the number is 11 digits, so positions 10 down to 0.

Wait, let me write the binary number as:

From left to right: positions 10 to 0:

Digit positions:

Position 10: 1

Position 9: 0

Position 8: 1

Position 7: 0

Position 6: 1

Position 5: 0

Position 4: 0

Position 3: 1

Position 2: 1

Position 1: 1

Position 0: 0

Wait, let me check:

Original binary number: 1 0 1 0 1 0 0 1 1 1 0

So leftmost digit is position 10 (since there are 11 digits), and rightmost is position 0.

So the value is:

1*(2^10) + 0*(2^9) + 1*(2^8) + 0*(2^7) + 1*(2^6) + 0*(2^5) + 0*(2^4) + 1*(2^3) + 1*(2^2) + 1*(2^1) + 0*(2^0)

Calculating each term:

2^10 = 1024

2^8 = 256

2^6 = 64

2^3 = 8

2^2 = 4

2^1 = 2

So adding them up:

1*1024 = 1024

0*512 = 0

1*256 = 256

0*128 = 0

1*64 = 64

0*32 = 0

0*16 = 0

1*8 = 8

1*4 = 4

1*2 = 2

0*1 = 0

Adding all these together:

1024 + 256 = 1280

1280 + 64 = 1344

1344 + 8 = 1352

1352 + 4 = 1356

1356 + 2 = 1358

So decimal value is 1358.

Now, converting 1358 decimal to octal.

To convert decimal to octal, divide by 8 and find remainders.

Let me do that:

1358 ÷ 8 = 169.75, so quotient 169, remainder 1358 - 169*8 = 1358 - 1352 = 6.

So first remainder (rightmost digit): 6.

169 ÷ 8 = 21.125, quotient 21, remainder 169 - 21*8 = 169 - 168 = 1.

Second remainder: 1.

21 ÷ 8 = 2.625, quotient 2, remainder 21 - 16 = 5.

Third remainder: 5.

2 ÷ 8 = 0.25, quotient 0, remainder 2.

Fourth remainder: 2.

So writing the remainders from last to first: 2, 5, 1, 6. So octal number is 2516₈. Which matches the previous result. So that's good. So the answer is 2516 in octal.

But let me check once more with another method. Maybe converting binary to octal directly by another grouping.

Alternatively, since each octal digit is three bits, maybe I can split the original binary number into groups of three, but without adding leading zeros. Wait, but the original number has 11 bits. If I group from the right, as we did before, adding one leading zero, then we get 2516. But maybe I can also split the original number into groups of three, starting from the right, and pad with leading zeros if necessary.

Original binary number: 10101001110

Split into groups of three from the right:

Starting from the right, the last three digits are 110, then the next three digits are 001, then the next three digits are 101, and then the remaining digits are 010. Wait, that's the same as before. So 010 101 001 110, which is 2 5 1 6. So same result.

Alternatively, if I didn't add the leading zero, but tried to group from the left, but that would be problematic because the leftmost group would have only two bits. So adding leading zeros is necessary.

Alternatively, maybe I can convert the binary number to hexadecimal first, then to octal? Let me try that as another check.

First, converting binary to hexadecimal. Since hexadecimal is base 16, which is 2⁴, so group into fours.

Original binary number: 10101001110. Let me count the digits again: 11 digits. To group into fours, need to add leading zeros to make it a multiple of four. 11 divided by 4 is 2 with remainder 3, so add one leading zero to make it 12 digits. Wait, 11 + 1 = 12, which is divisible by 4? No, 12 is divisible by 4? 12 / 4 = 3. Yes. So adding one leading zero: 0 1 0 1 0 1 0 0 1 1 1 0.

Now, grouping into fours from the left:

Group 1: 0101

Group 2: 0100

Group 3: 1110

Wait, let me check:

Original number with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of four:

Group 1: 0101

Group 2: 0100

Group 3: 1110

Wait, but that's three groups. Let me check the digits:

Group 1: first four digits: 0 1 0 1

Group 2: next four: 0 1 0 0

Group 3: next four: 1 1 1 0

Yes. Now converting each group to hexadecimal:

Group 1: 0101 is 5 in hex (since 0*8 + 1*4 + 0*2 + 1*1 = 5)

Group 2: 0100 is 4 in hex (0*8 + 1*4 + 0*2 + 0*1 = 4)

Group 3: 1110 is E in hex (1*8 + 1*4 + 1*2 + 0*1 = 14, which is E)

So hexadecimal number is 5 4 E, which is 54E₁₆.

Now, converting 54E₁₆ to octal. How to do that? Well, one way is to convert hexadecimal to decimal first, then decimal to octal. But we already know that 54E₁₆ is equal to 1358 decimal, which we found earlier. So converting 1358 decimal to octal gives 2516, which matches. Alternatively, maybe convert hexadecimal to octal directly?

But since hexadecimal is base 16 and octal is base 8, which are not powers of the same base, it's not straightforward. However, since 16 is 2⁴ and 8 is 2³, maybe we can convert each hexadecimal digit to 4 binary digits, then group into 3s for octal. Let me try that.

So 54E₁₆ is 0101 0100 1110 in binary. Wait, that's the same as before. Then, converting that to octal by grouping into 3s. Let me take the binary number 010101001110. Let me add leading zeros if necessary. Wait, the binary number is 12 bits: 010101001110. Let me group into 3s from the right:

Original binary: 010101001110

Split into groups of three from the right:

Starting from the right:

Group 1: 110

Group 2: 001

Group 3: 101

Group 4: 010

Which is the same as before. So converting each group:

110 = 6, 001 = 1, 101 = 5, 010 = 2. So octal number is 2516. Same result. So that's consistent.

Therefore, all methods lead to the octal number 2516. Therefore, I can be confident that the answer is 2516 in octal.

But just to make sure, let me check once again by converting the binary number to decimal and then to octal, as I did before.

We had the binary number 10101001110₂ equals 1358 decimal. Converting 1358 to octal:

Divide 1358 by 8:

1358 ÷ 8 = 169.75 → quotient 169, remainder 6

169 ÷ 8 = 21.125 → quotient 21, remainder 1

21 ÷ 8 = 2.625 → quotient 2, remainder 5

2 ÷ 8 = 0.25 → quotient 0, remainder 2

Reading the remainders from last to first: 2, 5, 1, 6 → 2516₈. Yep, same result.

Therefore, I think the answer is definitely 2516 in octal.

But just to make sure I didn't make any mistake in grouping, let me write the binary number again with the added leading zero and group them:

Original binary: 10101001110

Add leading zero: 010101001110

Split into groups of three from the right:

Starting from the right:

First group: 110 (positions 0-2)

Second group: 001 (positions 3-5)

Third group: 101 (positions 6-8)

Fourth group: 010 (positions 9-11)

Wait, positions are confusing, but if I write the binary number as:

0 1 0 1 0 1 0 0 1 1 1 0

Breaking into groups of three from the right:

Group 1: last three digits: 1 1 0 → 110

Group 2: previous three: 0 0 1 → 001

Group 3: previous three: 1 0 1 → 101

Group 4: remaining three: 0 1 0 → 010

So converting each group:

Group 1: 110 = 6

Group 2: 001 = 1

Group 3: 101 = 5

Group 4: 010 = 2

So octal digits are 2 (from group 4), 5 (group 3), 1 (group 2), 6 (group 1). So 2 5 1 6 → 2516. Correct.

Alternatively, if I write the groups in order from left to right, which is group 4, group 3, group 2, group 1. Wait, no. When grouping from the right, the first group is the rightmost, so when writing the octal number, the leftmost group is the leftmost group of the binary number. Wait, maybe I need to think of the groups as follows:

Original binary with leading zero: 0 1 0 1 0 1 0 0 1 1 1 0

Split into groups of three:

Group 1 (leftmost): 0 1 0

Group 2: 1 0 1

Group 3: 0 0 1

Group 4 (rightmost): 1 1 0

So converting each group:

Group 1: 010 = 2

Group 2: 101 = 5

Group 3: 001 = 1

Group 4: 110 = 6

So octal number is 2 5 1 6 → 2516. Same result. So regardless of the grouping direction, as long as I'm consistent, I get the same answer.

Therefore, I can be confident that the octal equivalent of the binary number 10101001110₂ is 2516₈.

**Final Answer**
The binary number $10101001110_{2}$ is equal to \boxed{2516} in base eight.

To convert the binary number $10101001110_2$ to its octal equivalent, we can use the fact that **each octal digit corresponds to exactly three binary digits**. This is because $8 = 2^3$, so we can group the binary digits into sets of three, starting from the right, and then convert each group to its octal equivalent.

---

### Step 1: Count the number of binary digits

The binary number is:

$$
10101001110
$$

This has **11 digits**. Since 11 is not a multiple of 3, we need to **add one leading zero** to make the total number of digits a multiple of 3:

$$
010101001110
$$

Now we have **12 digits**, which is divisible by 3.

---

### Step 2: Group the binary digits into sets of three

Starting from the **right**, we group the binary digits into sets of three:

$$
010\ 101\ 001\ 110
$$

---

### Step 3: Convert each group to octal

Now, convert each 3-bit binary group to its octal equivalent:

- $010_2 = 2_8$
- $101_2 = 5_8$
- $001_2 = 1_8$
- $110_2 = 6_8$

---

### Step 4: Combine the octal digits

Putting the octal digits together in the same order as the groups:

$$
2\ 5\ 1\ 6
$$

So, the octal representation is:

$$
\boxed{2516}
$$

---

### Final Answer

$$
\boxed{2516}
$$

Solution: Since $2^3=8$, we may convert between base 2 and base 8 representations by replacing each block of three digits in base 2 with its equivalent in base 8. In this case, we begin by noticing that the last three digits are worth $110_2=6_8$. The next block of three digits is $001_2=1_8$. Continuing, we find that the next two digits (moving right-to-left) are $101_2=5_8$ and $010_2=2_8$. Altogether, we find that $10101001110_{2}=\boxed{2516_8}$.


Explanation:
Okay, so I need to figure out the value of 9 & 2 using this operation they defined. Let me start by understanding the problem again. The operation & is defined for positive integers a and b as a & b equals the square root of (ab + a) divided by the square root of (ab - b). And I need to compute 9 & 2. Alright, let me write that down step by step.

First, let me recall the definition:

a & b = √(ab + a) / √(ab - b)

So, substituting a = 9 and b = 2 into this formula. Let me compute the numerator and the denominator separately.

Starting with the numerator: √(ab + a). Let me compute ab first. a is 9, b is 2, so ab is 9*2 = 18. Then add a, which is 9, so 18 + 9 = 27. Therefore, the numerator is √27.

Now the denominator: √(ab - b). Again, ab is 18, and then subtract b, which is 2. So 18 - 2 = 16. Therefore, the denominator is √16.

So putting it all together, 9 & 2 is √27 divided by √16. Let me write that as √27 / √16. Hmm, I can simplify this. Let me recall that √a / √b is equal to √(a/b). So maybe I can combine them into a single square root?

Alternatively, I can simplify each square root separately. Let me try that. √27 can be simplified. 27 is 9*3, so √27 = √(9*3) = √9 * √3 = 3√3. And √16 is 4. Therefore, the entire expression becomes 3√3 / 4. Is that the simplest radical form? Let me check.

Yes, 3√3 over 4 is a common fraction in simplest radical form. So is that the answer? Wait, let me verify my steps again to make sure I didn't make a mistake.

First, compute ab: 9*2 = 18. Then numerator inside the square root: ab + a = 18 + 9 = 27. Correct. Denominator inside the square root: ab - b = 18 - 2 = 16. Correct. Then √27 / √16. Simplify √27 to 3√3 and √16 is 4. So 3√3 / 4. That seems right.

Wait, but let me check if there's another way to approach this. Maybe rationalizing or something else? But since both numerator and denominator are square roots, and we simplified them, I think that's as simplified as it gets. Let me check if 27 and 16 have any common factors? Well, 27 is 3^3 and 16 is 2^4. So no common factors other than 1. Therefore, the fraction 3√3 / 4 is indeed in simplest form.

Alternatively, if I had combined the square roots first, √(27/16) which is √27 / √16, same thing. So that's the same as 3√3 / 4. So I think that's correct.

But let me check if I substituted the numbers correctly. The operation is a & b = √(ab + a)/√(ab - b). So for a=9, b=2, that's √(9*2 + 9)/√(9*2 - 2). Which is √(18 + 9)/√(18 - 2) = √27 / √16. Yep, that's exactly what I did. So that seems correct.

Wait, but let me check if there's a possibility that the operation is defined differently? For example, maybe the denominator is ab - b, but is there a chance that it's a different expression? The problem says "the operation & is defined for positive integers a and b as a & b = √(ab + a)/√(ab - b)". So no, that's exactly how it's defined. So I think my calculation is correct.

Alternatively, maybe I need to rationalize the denominator or something? But since the denominator is already a rational number (4), and the numerator is 3√3, there's no need to rationalize. So 3√3 over 4 is the simplest form.

Wait, but let me check if the problem says "common fraction in simplest radical form". So a common fraction would be a fraction where numerator and denominator are integers, and the radical is in the numerator. Which is exactly what I have here. So 3√3 over 4 is a common fraction (since 3 and 4 are integers) and it's in simplest radical form because √3 can't be simplified further. So that's the answer.

But just to be thorough, let me check if I made any arithmetic errors. Let me recompute ab + a and ab - b.

ab is 9*2 = 18. Then ab + a is 18 + 9 = 27. Correct. ab - b is 18 - 2 = 16. Correct. Then square roots: √27 is 3√3, √16 is 4. So 3√3 / 4. Yep. So I think that's correct.

Alternatively, maybe the problem expects something else? Let me think. For example, if there was a miscalculation in the expression. Let me check the original definition again. The operation is a & b equals sqrt(ab + a) divided by sqrt(ab - b). So that's sqrt(ab + a)/sqrt(ab - b). So that's exactly what I did. So no, I think that's correct.

Alternatively, maybe the problem is expecting me to write it as a single square root? Like sqrt( (ab + a)/(ab - b) ). Which would be sqrt(27/16) which is sqrt(27)/sqrt(16) which is same as 3√3 / 4. So same answer. So regardless of how I approach it, I get the same result.

Therefore, I think the answer is 3√3 over 4. Let me check if that's the simplest form. Since 3 and 4 have no common factors, and √3 is simplified, yes. So I think that's the answer.

Wait, but let me check if there's a different interpretation of the operation. For example, maybe the operation is defined as sqrt(ab + a) divided by sqrt(ab - b), but maybe there's some order of operations issue? Like, is it sqrt(ab + a) divided by sqrt(ab - b), or is it sqrt( (ab + a)/ (ab - b) )? But according to the original definition, it's written as [sqrt(ab + a)] / [sqrt(ab - b)], so that's the same as sqrt( (ab + a)/(ab - b) ). So both ways, same result. So that's correct.

Alternatively, maybe the problem is expecting me to rationalize the denominator? But in this case, the denominator is already rational. So 3√3 / 4 is fine. If the denominator was irrational, then we would need to rationalize, but here denominator is 4, which is rational. So no need. Therefore, 3√3 / 4 is the answer.

Therefore, I think I can confidently say that 9 & 2 is 3√3 over 4.

**Final Answer**
The value of $9 \& 2$ is \boxed{\dfrac{3\sqrt{3}}{4}}.

To evaluate the expression $ 9 \& 2 $, we begin by applying the given definition of the operation $ \& $:

$$
a \& b = \frac{\sqrt{ab + a}}{\sqrt{ab - b}}
$$

---

### Step 1: Substitute $ a = 9 $ and $ b = 2 $

We compute the numerator and denominator separately.

#### Numerator:
$$
ab + a = 9 \cdot 2 + 9 = 18 + 9 = 27
$$
$$
\sqrt{ab + a} = \sqrt{27}
$$

#### Denominator:
$$
ab - b = 9 \cdot 2 - 2 = 18 - 2 = 16
$$
$$
\sqrt{ab - b} = \sqrt{16}
$$

---

### Step 2: Simplify the Square Roots

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$
$$
\sqrt{16} = 4
$$

---

### Step 3: Combine the Results

$$
9 \& 2 = \frac{\sqrt{27}}{\sqrt{16}} = \frac{3\sqrt{3}}{4}
$$

This expression is already in **simplest radical form**, as the numerator contains a radical that cannot be simplified further, and the denominator is a rational number.

---

### Final Answer

$$
\boxed{\dfrac{3\sqrt{3}}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the value of 9 & 2 using this operation they defined. Let me start by understanding the problem again. The operation & is defined for positive integers a and b as a & b equals the square root of (ab + a) divided by the square root of (ab - b). And I need to compute 9 & 2. Alright, let me write that down step by step.

First, let me recall the definition:

a & b = √(ab + a) / √(ab - b)

So, substituting a = 9 and b = 2 into this formula. Let me compute the numerator and the denominator separately.

Starting with the numerator: √(ab + a). Let me compute ab first. a is 9, b is 2, so ab is 9*2 = 18. Then add a, which is 9, so 18 + 9 = 27. Therefore, the numerator is √27.

Now the denominator: √(ab - b). Again, ab is 18, and then subtract b, which is 2. So 18 - 2 = 16. Therefore, the denominator is √16.

So putting it all together, 9 & 2 is √27 divided by √16. Let me write that as √27 / √16. Hmm, I can simplify this. Let me recall that √a / √b is equal to √(a/b). So maybe I can combine them into a single square root?

Alternatively, I can simplify each square root separately. Let me try that. √27 can be simplified. 27 is 9*3, so √27 = √(9*3) = √9 * √3 = 3√3. And √16 is 4. Therefore, the entire expression becomes 3√3 / 4. Is that the simplest radical form? Let me check.

Yes, 3√3 over 4 is a common fraction in simplest radical form. So is that the answer? Wait, let me verify my steps again to make sure I didn't make a mistake.

First, compute ab: 9*2 = 18. Then numerator inside the square root: ab + a = 18 + 9 = 27. Correct. Denominator inside the square root: ab - b = 18 - 2 = 16. Correct. Then √27 / √16. Simplify √27 to 3√3 and √16 is 4. So 3√3 / 4. That seems right.

Wait, but let me check if there's another way to approach this. Maybe rationalizing or something else? But since both numerator and denominator are square roots, and we simplified them, I think that's as simplified as it gets. Let me check if 27 and 16 have any common factors? Well, 27 is 3^3 and 16 is 2^4. So no common factors other than 1. Therefore, the fraction 3√3 / 4 is indeed in simplest form.

Alternatively, if I had combined the square roots first, √(27/16) which is √27 / √16, same thing. So that's the same as 3√3 / 4. So I think that's correct.

But let me check if I substituted the numbers correctly. The operation is a & b = √(ab + a)/√(ab - b). So for a=9, b=2, that's √(9*2 + 9)/√(9*2 - 2). Which is √(18 + 9)/√(18 - 2) = √27 / √16. Yep, that's exactly what I did. So that seems correct.

Wait, but let me check if there's a possibility that the operation is defined differently? For example, maybe the denominator is ab - b, but is there a chance that it's a different expression? The problem says "the operation & is defined for positive integers a and b as a & b = √(ab + a)/√(ab - b)". So no, that's exactly how it's defined. So I think my calculation is correct.

Alternatively, maybe I need to rationalize the denominator or something? But since the denominator is already a rational number (4), and the numerator is 3√3, there's no need to rationalize. So 3√3 over 4 is the simplest form.

Wait, but let me check if the problem says "common fraction in simplest radical form". So a common fraction would be a fraction where numerator and denominator are integers, and the radical is in the numerator. Which is exactly what I have here. So 3√3 over 4 is a common fraction (since 3 and 4 are integers) and it's in simplest radical form because √3 can't be simplified further. So that's the answer.

But just to be thorough, let me check if I made any arithmetic errors. Let me recompute ab + a and ab - b.

ab is 9*2 = 18. Then ab + a is 18 + 9 = 27. Correct. ab - b is 18 - 2 = 16. Correct. Then square roots: √27 is 3√3, √16 is 4. So 3√3 / 4. Yep. So I think that's correct.

Alternatively, maybe the problem expects something else? Let me think. For example, if there was a miscalculation in the expression. Let me check the original definition again. The operation is a & b equals sqrt(ab + a) divided by sqrt(ab - b). So that's sqrt(ab + a)/sqrt(ab - b). So that's exactly what I did. So no, I think that's correct.

Alternatively, maybe the problem is expecting me to write it as a single square root? Like sqrt( (ab + a)/(ab - b) ). Which would be sqrt(27/16) which is sqrt(27)/sqrt(16) which is same as 3√3 / 4. So same answer. So regardless of how I approach it, I get the same result.

Therefore, I think the answer is 3√3 over 4. Let me check if that's the simplest form. Since 3 and 4 have no common factors, and √3 is simplified, yes. So I think that's the answer.

Wait, but let me check if there's a different interpretation of the operation. For example, maybe the operation is defined as sqrt(ab + a) divided by sqrt(ab - b), but maybe there's some order of operations issue? Like, is it sqrt(ab + a) divided by sqrt(ab - b), or is it sqrt( (ab + a)/ (ab - b) )? But according to the original definition, it's written as [sqrt(ab + a)] / [sqrt(ab - b)], so that's the same as sqrt( (ab + a)/(ab - b) ). So both ways, same result. So that's correct.

Alternatively, maybe the problem is expecting me to rationalize the denominator? But in this case, the denominator is already rational. So 3√3 / 4 is fine. If the denominator was irrational, then we would need to rationalize, but here denominator is 4, which is rational. So no need. Therefore, 3√3 / 4 is the answer.

Therefore, I think I can confidently say that 9 & 2 is 3√3 over 4.

**Final Answer**
The value of $9 \& 2$ is \boxed{\dfrac{3\sqrt{3}}{4}}.

To evaluate the expression $ 9 \& 2 $, we begin by applying the given definition of the operation $ \& $:

$$
a \& b = \frac{\sqrt{ab + a}}{\sqrt{ab - b}}
$$

---

### Step 1: Substitute $ a = 9 $ and $ b = 2 $

We compute the numerator and denominator separately.

#### Numerator:
$$
ab + a = 9 \cdot 2 + 9 = 18 + 9 = 27
$$
$$
\sqrt{ab + a} = \sqrt{27}
$$

#### Denominator:
$$
ab - b = 9 \cdot 2 - 2 = 18 - 2 = 16
$$
$$
\sqrt{ab - b} = \sqrt{16}
$$

---

### Step 2: Simplify the Square Roots

$$
\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3\sqrt{3}
$$
$$
\sqrt{16} = 4
$$

---

### Step 3: Combine the Results

$$
9 \& 2 = \frac{\sqrt{27}}{\sqrt{16}} = \frac{3\sqrt{3}}{4}
$$

This expression is already in **simplest radical form**, as the numerator contains a radical that cannot be simplified further, and the denominator is a rational number.

---

### Final Answer

$$
\boxed{\dfrac{3\sqrt{3}}{4}}
$$

Solution: We have $9\&2 = \frac{\sqrt{(9)(2)+9}}{\sqrt{(9)(2)-2}} = \frac{\sqrt{27}}{\sqrt{16}} = \boxed{\frac{3\sqrt{3}}{4}}.$


Explanation:
Okay, so I need to find the probability that the product of two numbers rolled on two fair 6-sided dice is a multiple of 5. Hmm, let me think about how to approach this.

First, I know that when dealing with probabilities involving dice, it's often helpful to consider the total number of possible outcomes and then figure out how many of those outcomes satisfy the condition. Since each die has 6 sides, the total number of possible outcomes when throwing two dice is 6 multiplied by 6, which is 36. So, there are 36 possible outcomes in total. Got that down.

Now, the question is about the product being a multiple of 5. Let me recall that a number is a multiple of 5 if it's divisible by 5. So, the product of the two numbers must be divisible by 5. To figure out how many of the 36 outcomes satisfy this, I need to determine in how many cases the product is divisible by 5.

Alternatively, maybe it's easier to calculate the probability of the complementary event and subtract it from 1. The complementary event here would be that the product is not a multiple of 5. That is, the product is not divisible by 5. So, if I can find the probability that the product is not divisible by 5, then subtracting that from 1 will give me the desired probability. Let me see if that approach is simpler.

To have the product not divisible by 5, neither of the numbers can be divisible by 5. Wait, is that correct? Let me think. If either of the numbers is divisible by 5, then the product will be divisible by 5. Conversely, if neither number is divisible by 5, then the product won't be divisible by 5. So yes, the complementary event is that neither die shows a 5. Therefore, the number of outcomes where the product is not divisible by 5 is equal to the number of outcomes where both dice show numbers that are not 5.

So, how many numbers on a die are not 5? Well, a die has numbers 1 through 6. So, excluding 5, there are 5 numbers: 1, 2, 3, 4, 6. Therefore, for each die, there are 5 possible outcomes that are not 5. Since the two dice are independent, the number of outcomes where neither die shows a 5 is 5 * 5 = 25. Therefore, the number of outcomes where the product is not divisible by 5 is 25. Therefore, the number of outcomes where the product is divisible by 5 is total outcomes minus these, which is 36 - 25 = 11. Wait, hold on, 36 - 25 is 11? Wait, 36 - 25 is 11? Wait, 25 + 11 is 36? Let me check: 25 + 10 is 35, plus 1 is 36. Yes, so 36 - 25 is 11. Therefore, there are 11 favorable outcomes. Therefore, the probability would be 11/36.

But wait, that seems low. Let me check again. Maybe I made a mistake here. Let me think again.

Alternatively, maybe I can approach it directly. For the product to be divisible by 5, at least one of the numbers must be divisible by 5. So, the number of outcomes where at least one die shows a 5.

So, how many outcomes have at least one 5? Well, that can be calculated as the number of outcomes where the first die is 5 plus the number of outcomes where the second die is 5 minus the number of outcomes where both dice are 5 (to correct for double-counting).

So, the number of outcomes where the first die is 5 is 1 * 6 = 6. Similarly, the number of outcomes where the second die is 5 is also 6. However, the overlap where both are 5 is 1 outcome. Therefore, the total number of favorable outcomes is 6 + 6 - 1 = 11. Which matches the previous result. Therefore, the probability is 11/36.

But wait, let me verify this with another approach. Let me list all possible outcomes where the product is divisible by 5.

First, the numbers on a die are 1, 2, 3, 4, 5, 6. So, for the product to be divisible by 5, at least one of the numbers must be 5. So, let's count how many such outcomes there are.

Case 1: First die is 5. Then, the second die can be anything from 1 to 6. So, that's 6 outcomes: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6).

Case 2: Second die is 5, but the first die is not 5. Because if the first die is already 5, we've already counted those in Case 1. So, the first die can be 1, 2, 3, 4, 6 (so 5 possibilities) and the second die is 5. Therefore, that's 5 outcomes: (1,5), (2,5), (3,5), (4,5), (6,5).

Therefore, total number of favorable outcomes is 6 + 5 = 11. Which again gives 11/36. So, that seems consistent.

But wait, let me check if there are any other cases. For example, if both dice are 5, that's already included in Case 1. So, no overlap issues. So, total is indeed 11. Therefore, the probability is 11/36.

But let me think again. Maybe I need to check if there are other numbers that can make the product divisible by 5. For example, if one die is 5 and the other is something else, or if both are 5. But since 5 is prime, the product will be divisible by 5 if and only if at least one of the numbers is divisible by 5. Since 5 is prime, so yes, that's correct. Therefore, the only way for the product to be divisible by 5 is if at least one die shows a 5. Therefore, the number of favorable outcomes is indeed 11.

But let me check with another approach. Let me think about the probability that the first die is not 5 and the second die is not 5. Which is (5/6)*(5/6) = 25/36. Therefore, the probability that at least one die is 5 is 1 - 25/36 = 11/36. Which again gives the same result.

Therefore, all approaches lead to 11/36. Therefore, the probability is 11/36.

But wait, just to be thorough, let me list all the possible outcomes where the product is divisible by 5.

First, when the first die is 5:

(5,1): product 5*1=5, which is divisible by 5.

(5,2): 10, divisible by 5.

(5,3): 15, divisible by 5.

(5,4): 20, divisible by 5.

(5,5): 25, divisible by 5.

(5,6): 30, divisible by 5.

That's 6 outcomes.

Then, when the second die is 5 and the first die is not 5:

(1,5): 5, divisible by 5.

(2,5): 10, divisible by 5.

(3,5): 15, divisible by 5.

(4,5): 20, divisible by 5.

(6,5): 30, divisible by 5.

That's 5 outcomes.

So total of 6 + 5 = 11 outcomes.

So, yes, that's 11 outcomes. Therefore, 11/36 is correct.

But let me check if there are any other outcomes where the product is divisible by 5. For example, if one die is 10? Wait, but the dice only go up to 6. So, the numbers on the dice are 1-6. Therefore, the only way to get a product divisible by 5 is to have at least one 5. Because 5 is prime, and the other numbers don't have 5 as a factor. For example, if you have a 5 and a 2, the product is 10, which is divisible by 5. If you have a 5 and a 3, it's 15, which is divisible by 5. Similarly, 5 and 4 is 20, 5 and 5 is 25, 5 and 6 is 30. All of these are divisible by 5.

If neither die is 5, then the product can't be divisible by 5. For example, if you have 2 and 3, product is 6, not divisible by 5. 2 and 4 is 8, not divisible by 5. 3 and 4 is 12, etc. So, indeed, only when at least one die is 5.

Therefore, the answer is 11/36.

But wait, let me check once more. Maybe I made a mistake in counting? Let me count the number of outcomes where the product is divisible by 5.

Total outcomes: 36.

Number of outcomes with at least one 5: 11.

Alternatively, let me think of the possible products.

But maybe that's more complicated. Alternatively, think of the possible pairs:

(1,5), (2,5), (3,5), (4,5), (5,5), (5,1), (5,2), (5,3), (5,4), (5,6), (6,5).

That's 11 pairs.

Yes, that's 11. So, that's correct.

Therefore, the probability is 11/36.

But wait, just to be 100% sure, let me check if there's a case where even if neither die is 5, the product is still divisible by 5. For example, if one die is 5 and the other is... no, if neither die is 5, then neither number is divisible by 5, so their product can't be divisible by 5. Because 5 is prime. So, if a number is not divisible by 5, then it's co-prime with 5. Therefore, the product of two numbers not divisible by 5 will also not be divisible by 5.

Therefore, yes, the only way for the product to be divisible by 5 is if at least one of the numbers is divisible by 5, which is only possible if at least one die shows a 5.

Therefore, the answer is 11/36.

But let me check with another method. Let me compute the probability that the first die is 5 or the second die is 5.

Probability that first die is 5: 1/6.

Probability that second die is 5: 1/6.

But the probability that either is 5 is P(A or B) = P(A) + P(B) - P(A and B).

Which is 1/6 + 1/6 - (1/6 * 1/6) = 2/6 - 1/36 = 12/36 - 1/36 = 11/36.

Same result.

Therefore, all methods converge to 11/36.

Therefore, I think that is the correct answer.

But just to be thorough, let me think if there's any other way this could be approached. For example, considering the possible products.

The possible products when rolling two dice range from 1*1=1 up to 6*6=36.

But enumerating all products and checking which are multiples of 5 would be tedious, but maybe for verification.

Multiples of 5 between 1 and 36 are 5, 10, 15, 20, 25, 30, 35.

So, let's see which products can result in these numbers.

First, 5: possible pairs are (1,5), (5,1).

10: (2,5), (5,2), (1,10) but 10 isn't on a die, so only (2,5), (5,2). Wait, also (5,2) and (2,5).

Wait, 10 can be achieved by (2,5), (5,2).

Similarly, 15: (3,5), (5,3).

20: (4,5), (5,4).

25: (5,5).

30: (5,6), (6,5).

35: (5,7) but 7 isn't on a die. So, no.

So, the products that are multiples of 5 are 5, 10, 15, 20, 25, 30.

Now, let's count the number of pairs that result in each of these products:

- 5: 2 outcomes (1,5) and (5,1)
- 10: 2 outcomes (2,5) and (5,2)
- 15: 2 outcomes (3,5) and (5,3)
- 20: 2 outcomes (4,5) and (5,4)
- 25: 1 outcome (5,5)
- 30: 2 outcomes (5,6) and (6,5)

Adding these up: 2 + 2 + 2 + 2 + 1 + 2 = 11.

Yes! That's 11 outcomes. Therefore, confirming again that there are 11 favorable outcomes.

Therefore, the probability is 11/36.

So, after multiple approaches and checks, I can be confident that the probability is 11/36.

**Final Answer**
The probability is \boxed{\dfrac{11}{36}}.

To determine the probability that the product of the numbers rolled on two fair 6-sided dice is a multiple of 5, we begin by analyzing the total number of possible outcomes and the conditions under which the product is divisible by 5.

---

### Step 1: Total Number of Outcomes

Each die has 6 faces, so when rolling two dice, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 2: Understanding the Condition

A number is divisible by 5 if it contains the prime factor 5. Since the product of two numbers is divisible by 5 **if and only if at least one of the numbers is divisible by 5**, we can reframe the problem as:

> What is the probability that **at least one** of the two dice shows a 5?

This is a classic case of the **complement rule**. Instead of directly counting the favorable outcomes, we can count the number of outcomes where **neither die shows a 5**, and subtract that from the total.

---

### Step 3: Complementary Event

If neither die shows a 5, then each die can only show one of the numbers: 1, 2, 3, 4, or 6 — that is, 5 possible outcomes per die.

So, the number of outcomes where **neither die shows a 5** is:

$$
5 \times 5 = 25
$$

Therefore, the number of outcomes where **at least one die shows a 5** is:

$$
36 - 25 = 11
$$

---

### Step 4: Final Probability

The probability that the product is a multiple of 5 is the number of favorable outcomes divided by the total number of outcomes:

$$
\frac{11}{36}
$$

---

### Step 5: Verification via Direct Counting

To confirm, we can list all the outcomes where the product is divisible by 5:

- (1,5), (5,1)
- (2,5), (5,2)
- (3,5), (5,3)
- (4,5), (5,4)
- (5,5)
- (5,6), (6,5)

That's a total of **11** outcomes, confirming our earlier result.

---

### Final Answer

$$
\boxed{\dfrac{11}{36}}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that the product of two numbers rolled on two fair 6-sided dice is a multiple of 5. Hmm, let me think about how to approach this.

First, I know that when dealing with probabilities involving dice, it's often helpful to consider the total number of possible outcomes and then figure out how many of those outcomes satisfy the condition. Since each die has 6 sides, the total number of possible outcomes when throwing two dice is 6 multiplied by 6, which is 36. So, there are 36 possible outcomes in total. Got that down.

Now, the question is about the product being a multiple of 5. Let me recall that a number is a multiple of 5 if it's divisible by 5. So, the product of the two numbers must be divisible by 5. To figure out how many of the 36 outcomes satisfy this, I need to determine in how many cases the product is divisible by 5.

Alternatively, maybe it's easier to calculate the probability of the complementary event and subtract it from 1. The complementary event here would be that the product is not a multiple of 5. That is, the product is not divisible by 5. So, if I can find the probability that the product is not divisible by 5, then subtracting that from 1 will give me the desired probability. Let me see if that approach is simpler.

To have the product not divisible by 5, neither of the numbers can be divisible by 5. Wait, is that correct? Let me think. If either of the numbers is divisible by 5, then the product will be divisible by 5. Conversely, if neither number is divisible by 5, then the product won't be divisible by 5. So yes, the complementary event is that neither die shows a 5. Therefore, the number of outcomes where the product is not divisible by 5 is equal to the number of outcomes where both dice show numbers that are not 5.

So, how many numbers on a die are not 5? Well, a die has numbers 1 through 6. So, excluding 5, there are 5 numbers: 1, 2, 3, 4, 6. Therefore, for each die, there are 5 possible outcomes that are not 5. Since the two dice are independent, the number of outcomes where neither die shows a 5 is 5 * 5 = 25. Therefore, the number of outcomes where the product is not divisible by 5 is 25. Therefore, the number of outcomes where the product is divisible by 5 is total outcomes minus these, which is 36 - 25 = 11. Wait, hold on, 36 - 25 is 11? Wait, 36 - 25 is 11? Wait, 25 + 11 is 36? Let me check: 25 + 10 is 35, plus 1 is 36. Yes, so 36 - 25 is 11. Therefore, there are 11 favorable outcomes. Therefore, the probability would be 11/36.

But wait, that seems low. Let me check again. Maybe I made a mistake here. Let me think again.

Alternatively, maybe I can approach it directly. For the product to be divisible by 5, at least one of the numbers must be divisible by 5. So, the number of outcomes where at least one die shows a 5.

So, how many outcomes have at least one 5? Well, that can be calculated as the number of outcomes where the first die is 5 plus the number of outcomes where the second die is 5 minus the number of outcomes where both dice are 5 (to correct for double-counting).

So, the number of outcomes where the first die is 5 is 1 * 6 = 6. Similarly, the number of outcomes where the second die is 5 is also 6. However, the overlap where both are 5 is 1 outcome. Therefore, the total number of favorable outcomes is 6 + 6 - 1 = 11. Which matches the previous result. Therefore, the probability is 11/36.

But wait, let me verify this with another approach. Let me list all possible outcomes where the product is divisible by 5.

First, the numbers on a die are 1, 2, 3, 4, 5, 6. So, for the product to be divisible by 5, at least one of the numbers must be 5. So, let's count how many such outcomes there are.

Case 1: First die is 5. Then, the second die can be anything from 1 to 6. So, that's 6 outcomes: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6).

Case 2: Second die is 5, but the first die is not 5. Because if the first die is already 5, we've already counted those in Case 1. So, the first die can be 1, 2, 3, 4, 6 (so 5 possibilities) and the second die is 5. Therefore, that's 5 outcomes: (1,5), (2,5), (3,5), (4,5), (6,5).

Therefore, total number of favorable outcomes is 6 + 5 = 11. Which again gives 11/36. So, that seems consistent.

But wait, let me check if there are any other cases. For example, if both dice are 5, that's already included in Case 1. So, no overlap issues. So, total is indeed 11. Therefore, the probability is 11/36.

But let me think again. Maybe I need to check if there are other numbers that can make the product divisible by 5. For example, if one die is 5 and the other is something else, or if both are 5. But since 5 is prime, the product will be divisible by 5 if and only if at least one of the numbers is divisible by 5. Since 5 is prime, so yes, that's correct. Therefore, the only way for the product to be divisible by 5 is if at least one die shows a 5. Therefore, the number of favorable outcomes is indeed 11.

But let me check with another approach. Let me think about the probability that the first die is not 5 and the second die is not 5. Which is (5/6)*(5/6) = 25/36. Therefore, the probability that at least one die is 5 is 1 - 25/36 = 11/36. Which again gives the same result.

Therefore, all approaches lead to 11/36. Therefore, the probability is 11/36.

But wait, just to be thorough, let me list all the possible outcomes where the product is divisible by 5.

First, when the first die is 5:

(5,1): product 5*1=5, which is divisible by 5.

(5,2): 10, divisible by 5.

(5,3): 15, divisible by 5.

(5,4): 20, divisible by 5.

(5,5): 25, divisible by 5.

(5,6): 30, divisible by 5.

That's 6 outcomes.

Then, when the second die is 5 and the first die is not 5:

(1,5): 5, divisible by 5.

(2,5): 10, divisible by 5.

(3,5): 15, divisible by 5.

(4,5): 20, divisible by 5.

(6,5): 30, divisible by 5.

That's 5 outcomes.

So total of 6 + 5 = 11 outcomes.

So, yes, that's 11 outcomes. Therefore, 11/36 is correct.

But let me check if there are any other outcomes where the product is divisible by 5. For example, if one die is 10? Wait, but the dice only go up to 6. So, the numbers on the dice are 1-6. Therefore, the only way to get a product divisible by 5 is to have at least one 5. Because 5 is prime, and the other numbers don't have 5 as a factor. For example, if you have a 5 and a 2, the product is 10, which is divisible by 5. If you have a 5 and a 3, it's 15, which is divisible by 5. Similarly, 5 and 4 is 20, 5 and 5 is 25, 5 and 6 is 30. All of these are divisible by 5.

If neither die is 5, then the product can't be divisible by 5. For example, if you have 2 and 3, product is 6, not divisible by 5. 2 and 4 is 8, not divisible by 5. 3 and 4 is 12, etc. So, indeed, only when at least one die is 5.

Therefore, the answer is 11/36.

But wait, let me check once more. Maybe I made a mistake in counting? Let me count the number of outcomes where the product is divisible by 5.

Total outcomes: 36.

Number of outcomes with at least one 5: 11.

Alternatively, let me think of the possible products.

But maybe that's more complicated. Alternatively, think of the possible pairs:

(1,5), (2,5), (3,5), (4,5), (5,5), (5,1), (5,2), (5,3), (5,4), (5,6), (6,5).

That's 11 pairs.

Yes, that's 11. So, that's correct.

Therefore, the probability is 11/36.

But wait, just to be 100% sure, let me check if there's a case where even if neither die is 5, the product is still divisible by 5. For example, if one die is 5 and the other is... no, if neither die is 5, then neither number is divisible by 5, so their product can't be divisible by 5. Because 5 is prime. So, if a number is not divisible by 5, then it's co-prime with 5. Therefore, the product of two numbers not divisible by 5 will also not be divisible by 5.

Therefore, yes, the only way for the product to be divisible by 5 is if at least one of the numbers is divisible by 5, which is only possible if at least one die shows a 5.

Therefore, the answer is 11/36.

But let me check with another method. Let me compute the probability that the first die is 5 or the second die is 5.

Probability that first die is 5: 1/6.

Probability that second die is 5: 1/6.

But the probability that either is 5 is P(A or B) = P(A) + P(B) - P(A and B).

Which is 1/6 + 1/6 - (1/6 * 1/6) = 2/6 - 1/36 = 12/36 - 1/36 = 11/36.

Same result.

Therefore, all methods converge to 11/36.

Therefore, I think that is the correct answer.

But just to be thorough, let me think if there's any other way this could be approached. For example, considering the possible products.

The possible products when rolling two dice range from 1*1=1 up to 6*6=36.

But enumerating all products and checking which are multiples of 5 would be tedious, but maybe for verification.

Multiples of 5 between 1 and 36 are 5, 10, 15, 20, 25, 30, 35.

So, let's see which products can result in these numbers.

First, 5: possible pairs are (1,5), (5,1).

10: (2,5), (5,2), (1,10) but 10 isn't on a die, so only (2,5), (5,2). Wait, also (5,2) and (2,5).

Wait, 10 can be achieved by (2,5), (5,2).

Similarly, 15: (3,5), (5,3).

20: (4,5), (5,4).

25: (5,5).

30: (5,6), (6,5).

35: (5,7) but 7 isn't on a die. So, no.

So, the products that are multiples of 5 are 5, 10, 15, 20, 25, 30.

Now, let's count the number of pairs that result in each of these products:

- 5: 2 outcomes (1,5) and (5,1)
- 10: 2 outcomes (2,5) and (5,2)
- 15: 2 outcomes (3,5) and (5,3)
- 20: 2 outcomes (4,5) and (5,4)
- 25: 1 outcome (5,5)
- 30: 2 outcomes (5,6) and (6,5)

Adding these up: 2 + 2 + 2 + 2 + 1 + 2 = 11.

Yes! That's 11 outcomes. Therefore, confirming again that there are 11 favorable outcomes.

Therefore, the probability is 11/36.

So, after multiple approaches and checks, I can be confident that the probability is 11/36.

**Final Answer**
The probability is \boxed{\dfrac{11}{36}}.

To determine the probability that the product of the numbers rolled on two fair 6-sided dice is a multiple of 5, we begin by analyzing the total number of possible outcomes and the conditions under which the product is divisible by 5.

---

### Step 1: Total Number of Outcomes

Each die has 6 faces, so when rolling two dice, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 2: Understanding the Condition

A number is divisible by 5 if it contains the prime factor 5. Since the product of two numbers is divisible by 5 **if and only if at least one of the numbers is divisible by 5**, we can reframe the problem as:

> What is the probability that **at least one** of the two dice shows a 5?

This is a classic case of the **complement rule**. Instead of directly counting the favorable outcomes, we can count the number of outcomes where **neither die shows a 5**, and subtract that from the total.

---

### Step 3: Complementary Event

If neither die shows a 5, then each die can only show one of the numbers: 1, 2, 3, 4, or 6 — that is, 5 possible outcomes per die.

So, the number of outcomes where **neither die shows a 5** is:

$$
5 \times 5 = 25
$$

Therefore, the number of outcomes where **at least one die shows a 5** is:

$$
36 - 25 = 11
$$

---

### Step 4: Final Probability

The probability that the product is a multiple of 5 is the number of favorable outcomes divided by the total number of outcomes:

$$
\frac{11}{36}
$$

---

### Step 5: Verification via Direct Counting

To confirm, we can list all the outcomes where the product is divisible by 5:

- (1,5), (5,1)
- (2,5), (5,2)
- (3,5), (5,3)
- (4,5), (5,4)
- (5,5)
- (5,6), (6,5)

That's a total of **11** outcomes, confirming our earlier result.

---

### Final Answer

$$
\boxed{\dfrac{11}{36}}
$$

Solution: Probability questions are sometimes answered by calculating the ways the event will NOT happen, then subtracting.  In this problem the $1$, $2$, $3$, $4$ and $6$ faces are paired to create $5 \times 5 = 25$ number pairs whose product is NOT multiples of 5.  This leaves $36 - 25 = 11$ ways to get a multiple of $5$, so the probability is $\boxed{\frac{11}{36}}$.


Explanation:
Okay, so I need to find real numbers a and b, where a isn't zero, such that the equation f(x) - f(a) equals (x - a)^2 times (x - b). The function f(x) is given as x³ + 3x² + 1. Alright, let me start by understanding what this equation is saying.

First, f(x) - f(a) is the difference between the function evaluated at x and at a. The right side is a cubic polynomial, specifically (x - a) squared times (x - b). So, expanding the right side should give me a cubic polynomial, and the left side is also a cubic polynomial. Therefore, if I can expand both sides and equate coefficients, maybe I can solve for a and b.

Let me write down both sides.

Left side: f(x) - f(a) = [x³ + 3x² + 1] - [a³ + 3a² + 1] = x³ + 3x² + 1 - a³ - 3a² - 1 = x³ + 3x² - a³ - 3a².

Simplify that: x³ + 3x² - (a³ + 3a²). So that's the left side.

Right side: (x - a)²(x - b). Let me expand that. First, expand (x - a)²: that's x² - 2a x + a². Then multiply by (x - b):

Multiply x² - 2a x + a² by x - b:

First, x²*(x - b) = x³ - b x².

Then, -2a x*(x - b) = -2a x² + 2a b x.

Then, a²*(x - b) = a² x - a² b.

So adding all those terms together:

x³ - b x² - 2a x² + 2a b x + a² x - a² b.

Combine like terms:

x³ + (-b - 2a) x² + (2a b + a²) x - a² b.

So the right side is x³ + (-b - 2a)x² + (2ab + a²)x - a² b.

Now, the left side is x³ + 3x² - (a³ + 3a²). Let me write that as x³ + 3x² + 0x - (a³ + 3a²). So comparing coefficients with the right side.

So equate coefficients for each power of x:

For x³: 1 = 1. That's already equal, so no problem.

For x²: 3 = -b - 2a. So equation (1): -b - 2a = 3.

For x term: 0 = 2ab + a². Equation (2): 2ab + a² = 0.

For the constant term: - (a³ + 3a²) = -a² b. Let me write that as equation (3): -a³ - 3a² = -a² b. Which simplifies to a³ + 3a² = a² b. Assuming a ≠ 0, we can divide both sides by a²: a + 3 = b. So equation (3) gives b = a + 3.

So now, from equation (3), we have b = a + 3. Let me substitute this into equation (1) and equation (2).

First, equation (1): -b - 2a = 3. Substitute b = a + 3:

- (a + 3) - 2a = 3 => -a - 3 - 2a = 3 => -3a - 3 = 3 => -3a = 6 => a = -2.

So a is -2. Then, from equation (3), b = a + 3 = -2 + 3 = 1. So b is 1. Let me check if this satisfies equation (2).

Equation (2): 2ab + a² = 0. Let's plug in a = -2, b = 1:

2*(-2)*(1) + (-2)^2 = -4 + 4 = 0. Perfect, that works. So that's consistent.

Therefore, the ordered pair (a, b) is (-2, 1). Let me just verify this with the original equation to make sure.

Let me compute f(x) - f(a) where a = -2. First, compute f(-2):

f(-2) = (-2)^3 + 3*(-2)^2 + 1 = -8 + 3*(4) + 1 = -8 + 12 + 1 = 5.

So f(x) - f(-2) = x³ + 3x² + 1 - 5 = x³ + 3x² - 4.

Now, compute (x - a)^2 (x - b) with a = -2 and b = 1: (x - (-2))² (x - 1) = (x + 2)^2 (x - 1). Let me expand that:

First, (x + 2)^2 = x² + 4x + 4. Multiply by (x - 1):

(x² + 4x + 4)(x - 1) = x³ - x² + 4x² - 4x + 4x - 4 = x³ + 3x² - 4. Which matches f(x) - f(-2). Perfect, so that works.

Therefore, the answer is (-2, 1). So the ordered pair is (-2, 1). Let me check if there are other possible solutions, but since we had equations that led us uniquely to a = -2 and b = 1, and the problem states that such a and b exist with a ≠ 0, which is satisfied here, so this must be the solution.

**Final Answer**
The ordered pair is \boxed{(-2, 1)}.

To solve the equation

$$
f(x) - f(a) = (x - a)^2 (x - b)
$$

where $ f(x) = x^3 + 3x^2 + 1 $, and given that $ a \neq 0 $, we aim to find the ordered pair $ (a, b) $.

---

### Step 1: Expand the Left-Hand Side

We compute $ f(x) - f(a) $:

$$
f(x) - f(a) = (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - a^3 - 3a^2
$$

So the left-hand side becomes:

$$
x^3 + 3x^2 - a^3 - 3a^2
$$

---

### Step 2: Expand the Right-Hand Side

We expand $ (x - a)^2 (x - b) $:

First, expand $ (x - a)^2 = x^2 - 2ax + a^2 $

Then multiply by $ (x - b) $:

$$
(x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b
$$

Group like terms:

$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

---

### Step 3: Equate Coefficients

Now equate the left-hand side and the right-hand side:

Left side:
$$
x^3 + 3x^2 + 0x - (a^3 + 3a^2)
$$

Right side:
$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

Matching coefficients:

1. $ x^3 $: $ 1 = 1 $ → OK
2. $ x^2 $: $ 3 = -b - 2a $ → Equation (1): $ -b - 2a = 3 $
3. $ x $: $ 0 = 2ab + a^2 $ → Equation (2): $ 2ab + a^2 = 0 $
4. Constant term: $ -a^3 - 3a^2 = -a^2b $ → Equation (3): $ a^3 + 3a^2 = a^2b $

---

### Step 4: Solve the System of Equations

From Equation (3):

$$
a^3 + 3a^2 = a^2b \Rightarrow b = a + 3 \quad \text{(since } a \neq 0 \text{)}
$$

Substitute $ b = a + 3 $ into Equation (1):

$$
-(a + 3) - 2a = 3 \Rightarrow -a - 3 - 2a = 3 \Rightarrow -3a - 3 = 3 \Rightarrow -3a = 6 \Rightarrow a = -2
$$

Now find $ b $:

$$
b = a + 3 = -2 + 3 = 1
$$

Check Equation (2):

$$
2ab + a^2 = 2(-2)(1) + (-2)^2 = -4 + 4 = 0 \quad \text{✓}
$$

---

### Step 5: Final Verification

Let’s verify the original equation with $ a = -2 $, $ b = 1 $:

- $ f(x) - f(-2) = x^3 + 3x^2 + 1 - f(-2) $
- $ f(-2) = (-2)^3 + 3(-2)^2 + 1 = -8 + 12 + 1 = 5 $
- So $ f(x) - f(-2) = x^3 + 3x^2 - 4 $

Now compute $ (x + 2)^2(x - 1) $:

- $ (x + 2)^2 = x^2 + 4x + 4 $
- Multiply by $ (x - 1) $:
  $$
  (x^2 + 4x + 4)(x - 1) = x^3 - x^2 + 4x^2 - 4x + 4x - 4 = x^3 + 3x^2 - 4
  $$

This matches the left-hand side, confirming the solution.

---

### Final Answer

$$
\boxed{(-2, 1)}
$$

[Assistant]
Explanation:
Okay, so I need to find real numbers a and b, where a isn't zero, such that the equation f(x) - f(a) equals (x - a)^2 times (x - b). The function f(x) is given as x³ + 3x² + 1. Alright, let me start by understanding what this equation is saying.

First, f(x) - f(a) is the difference between the function evaluated at x and at a. The right side is a cubic polynomial, specifically (x - a) squared times (x - b). So, expanding the right side should give me a cubic polynomial, and the left side is also a cubic polynomial. Therefore, if I can expand both sides and equate coefficients, maybe I can solve for a and b.

Let me write down both sides.

Left side: f(x) - f(a) = [x³ + 3x² + 1] - [a³ + 3a² + 1] = x³ + 3x² + 1 - a³ - 3a² - 1 = x³ + 3x² - a³ - 3a².

Simplify that: x³ + 3x² - (a³ + 3a²). So that's the left side.

Right side: (x - a)²(x - b). Let me expand that. First, expand (x - a)²: that's x² - 2a x + a². Then multiply by (x - b):

Multiply x² - 2a x + a² by x - b:

First, x²*(x - b) = x³ - b x².

Then, -2a x*(x - b) = -2a x² + 2a b x.

Then, a²*(x - b) = a² x - a² b.

So adding all those terms together:

x³ - b x² - 2a x² + 2a b x + a² x - a² b.

Combine like terms:

x³ + (-b - 2a) x² + (2a b + a²) x - a² b.

So the right side is x³ + (-b - 2a)x² + (2ab + a²)x - a² b.

Now, the left side is x³ + 3x² - (a³ + 3a²). Let me write that as x³ + 3x² + 0x - (a³ + 3a²). So comparing coefficients with the right side.

So equate coefficients for each power of x:

For x³: 1 = 1. That's already equal, so no problem.

For x²: 3 = -b - 2a. So equation (1): -b - 2a = 3.

For x term: 0 = 2ab + a². Equation (2): 2ab + a² = 0.

For the constant term: - (a³ + 3a²) = -a² b. Let me write that as equation (3): -a³ - 3a² = -a² b. Which simplifies to a³ + 3a² = a² b. Assuming a ≠ 0, we can divide both sides by a²: a + 3 = b. So equation (3) gives b = a + 3.

So now, from equation (3), we have b = a + 3. Let me substitute this into equation (1) and equation (2).

First, equation (1): -b - 2a = 3. Substitute b = a + 3:

- (a + 3) - 2a = 3 => -a - 3 - 2a = 3 => -3a - 3 = 3 => -3a = 6 => a = -2.

So a is -2. Then, from equation (3), b = a + 3 = -2 + 3 = 1. So b is 1. Let me check if this satisfies equation (2).

Equation (2): 2ab + a² = 0. Let's plug in a = -2, b = 1:

2*(-2)*(1) + (-2)^2 = -4 + 4 = 0. Perfect, that works. So that's consistent.

Therefore, the ordered pair (a, b) is (-2, 1). Let me just verify this with the original equation to make sure.

Let me compute f(x) - f(a) where a = -2. First, compute f(-2):

f(-2) = (-2)^3 + 3*(-2)^2 + 1 = -8 + 3*(4) + 1 = -8 + 12 + 1 = 5.

So f(x) - f(-2) = x³ + 3x² + 1 - 5 = x³ + 3x² - 4.

Now, compute (x - a)^2 (x - b) with a = -2 and b = 1: (x - (-2))² (x - 1) = (x + 2)^2 (x - 1). Let me expand that:

First, (x + 2)^2 = x² + 4x + 4. Multiply by (x - 1):

(x² + 4x + 4)(x - 1) = x³ - x² + 4x² - 4x + 4x - 4 = x³ + 3x² - 4. Which matches f(x) - f(-2). Perfect, so that works.

Therefore, the answer is (-2, 1). So the ordered pair is (-2, 1). Let me check if there are other possible solutions, but since we had equations that led us uniquely to a = -2 and b = 1, and the problem states that such a and b exist with a ≠ 0, which is satisfied here, so this must be the solution.

**Final Answer**
The ordered pair is \boxed{(-2, 1)}.

To solve the equation

$$
f(x) - f(a) = (x - a)^2 (x - b)
$$

where $ f(x) = x^3 + 3x^2 + 1 $, and given that $ a \neq 0 $, we aim to find the ordered pair $ (a, b) $.

---

### Step 1: Expand the Left-Hand Side

We compute $ f(x) - f(a) $:

$$
f(x) - f(a) = (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - a^3 - 3a^2
$$

So the left-hand side becomes:

$$
x^3 + 3x^2 - a^3 - 3a^2
$$

---

### Step 2: Expand the Right-Hand Side

We expand $ (x - a)^2 (x - b) $:

First, expand $ (x - a)^2 = x^2 - 2ax + a^2 $

Then multiply by $ (x - b) $:

$$
(x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b
$$

Group like terms:

$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

---

### Step 3: Equate Coefficients

Now equate the left-hand side and the right-hand side:

Left side:
$$
x^3 + 3x^2 + 0x - (a^3 + 3a^2)
$$

Right side:
$$
x^3 + (-b - 2a)x^2 + (2ab + a^2)x - a^2b
$$

Matching coefficients:

1. $ x^3 $: $ 1 = 1 $ → OK
2. $ x^2 $: $ 3 = -b - 2a $ → Equation (1): $ -b - 2a = 3 $
3. $ x $: $ 0 = 2ab + a^2 $ → Equation (2): $ 2ab + a^2 = 0 $
4. Constant term: $ -a^3 - 3a^2 = -a^2b $ → Equation (3): $ a^3 + 3a^2 = a^2b $

---

### Step 4: Solve the System of Equations

From Equation (3):

$$
a^3 + 3a^2 = a^2b \Rightarrow b = a + 3 \quad \text{(since } a \neq 0 \text{)}
$$

Substitute $ b = a + 3 $ into Equation (1):

$$
-(a + 3) - 2a = 3 \Rightarrow -a - 3 - 2a = 3 \Rightarrow -3a - 3 = 3 \Rightarrow -3a = 6 \Rightarrow a = -2
$$

Now find $ b $:

$$
b = a + 3 = -2 + 3 = 1
$$

Check Equation (2):

$$
2ab + a^2 = 2(-2)(1) + (-2)^2 = -4 + 4 = 0 \quad \text{✓}
$$

---

### Step 5: Final Verification

Let’s verify the original equation with $ a = -2 $, $ b = 1 $:

- $ f(x) - f(-2) = x^3 + 3x^2 + 1 - f(-2) $
- $ f(-2) = (-2)^3 + 3(-2)^2 + 1 = -8 + 12 + 1 = 5 $
- So $ f(x) - f(-2) = x^3 + 3x^2 - 4 $

Now compute $ (x + 2)^2(x - 1) $:

- $ (x + 2)^2 = x^2 + 4x + 4 $
- Multiply by $ (x - 1) $:
  $$
  (x^2 + 4x + 4)(x - 1) = x^3 - x^2 + 4x^2 - 4x + 4x - 4 = x^3 + 3x^2 - 4
  $$

This matches the left-hand side, confirming the solution.

---

### Final Answer

$$
\boxed{(-2, 1)}
$$

Solution:
By the remainder theorem, $f(x) - f(a)$ is divisible by $x - a,$ so we can take out a factor of $x - a$ accordingly:
\begin{align*}
f(x) - f(a) &= (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) \\
&= (x^3 - a^3) + 3(x^2 - a^2) \\
&= (x - a)(x^2 + ax + a^2) + 3(x - a)(x + a) \\
&= (x - a)(x^2 + ax + a^2 + 3x + 3a) \\
&= (x - a)(x^2 + (a + 3) x + a^2 + 3a).
\end{align*}Thus, we want
\[x^2 + (a + 3) x + a^2 + 3a = (x - a)(x - b) = x^2 - (a + b) x + ab.\]Matching coefficients, we get
\begin{align*}
a + 3 &= -a - b, \\
a^2 + 3a &= ab.
\end{align*}Since $a \neq 0,$ we can divide both sides of the second equation by $a,$ to get $a + 3 = b.$  Then $-a - b = b,$ so $a = -2b.$  Then
\[-2b + 3 = 2b - b,\]which gives us $b = 1$.  Then $a = -2,$ so $(a,b) = \boxed{(-2,1)}.$