Explanation:
Okay, so I need to find the remainder when this big polynomial is divided by x + 2. The polynomial is (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Hmm, dividing by x + 2... I remember that when you divide a polynomial by a linear divisor like x - a, the remainder is just the value of the polynomial at x = a. Wait, but here it's x + 2, which is the same as x - (-2). So, according to the Remainder Theorem, the remainder should be the value of the polynomial evaluated at x = -2. Is that right? Let me confirm. Yes, the Remainder Theorem says that if you divide a polynomial f(x) by x - a, the remainder is f(a). So in this case, since we're dividing by x + 2, which is x - (-2), the remainder is f(-2). So I just need to compute f(-2) where f(x) is that big expression. Alright, let me write that down.

So f(x) = (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Therefore, f(-2) = [5*(-2) + 9]^611 + [(-2) + 5]^11 + [(-2) - 1]^11 + 3*(-2)^2 + 1. Let me compute each term step by step.

First term: 5*(-2) + 9. Let me calculate that. 5*(-2) is -10, then adding 9 gives -10 + 9 = -1. So the first term is (-1)^611. Now, since 611 is an odd exponent, (-1)^odd is -1. So first term is -1.

Second term: (-2) + 5. That's 3. So the second term is 3^11. Let me compute 3^11. Hmm, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729, 3^7 = 2187, 3^8 = 6561, 3^9 = 19683, 3^10 = 59049, 3^11 = 177147. So that's 177,147. Let me check if I can compute that another way, but maybe I can just keep it as 3^11 for now. Wait, maybe I need the exact value? Or maybe there's a simplification? Let me see.

Third term: (-2) - 1 = -3. So the third term is (-3)^11. Again, exponent is odd, so that's -3^11. Which is -177,147. Wait, same as the second term but negative. So second term is +3^11, third term is -3^11. So they might cancel each other? Let me check. If I add them together, 3^11 - 3^11 = 0? Wait, no. Wait, the second term is [(-2) + 5]^11 = 3^11, and the third term is [(-2) - 1]^11 = (-3)^11 = -3^11. So adding them together, 3^11 + (-3^11) = 0. Oh! So those two terms cancel each other out. That's nice. So the second and third terms sum to zero. That simplifies things.

So then, moving on. The first term is -1, and then the rest is 0 plus the other terms. Let me check the rest. The fourth term is 3*(-2)^2 + 1. Let me compute that. (-2)^2 is 4, so 3*4 is 12, plus 1 is 13. So altogether, f(-2) is -1 + 0 + 13. Which is 12. Therefore, the remainder is 12? Wait, let me check again.

Wait, let me retrace. First term: (-1)^611 = -1. Second term: 3^11. Third term: (-3)^11 = -3^11. So second + third term: 3^11 - 3^11 = 0. Then fourth term: 3*(-2)^2 + 1. (-2)^2 is 4, so 3*4 is 12, plus 1 is 13. So total is -1 + 0 + 13 = 12. So the remainder is 12? That seems straightforward. Wait, but let me check if I did all steps correctly.

First term: 5x + 9 at x = -2: 5*(-2) + 9 = -10 + 9 = -1. Correct. Then exponent 611: (-1)^611 = -1. Correct. Second term: x + 5 at x = -2: -2 + 5 = 3. Correct. Third term: x - 1 at x = -2: -2 -1 = -3. Correct. Then exponents: 3^11 and (-3)^11. Correct. Then 3^11 - 3^11 = 0. Then the rest: 3x² + 1. At x = -2: 3*(4) + 1 = 12 + 1 = 13. Correct. Then total: -1 + 13 = 12. So the remainder is 12? Hmm. Wait, but let me check if there's any mistake in the exponents or signs.

Wait, the original polynomial is (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. So when I plug in x = -2, it's [(-1)^611] + [3^11] + [(-3)^11] + 3*(-2)^2 + 1. Which is -1 + 3^11 - 3^11 + 12 + 1. So yes, 3^11 - 3^11 cancels, then -1 + 12 + 1 = 12. So 12. So the remainder is 12? That seems correct. Wait, but let me check if I didn't make a mistake in the exponents. For example, is the exponent on the third term 11? Yes, the original problem says (x - 1)^11. So yes. So (-3)^11 is indeed -3^11. So that's correct.

Alternatively, maybe I can check with smaller exponents? Let me try with a smaller exponent to see if my approach is correct. Suppose instead of 611, I had (5x + 9)^1. Then f(-2) would be (-1)^1 + 3^1 + (-3)^1 + 3*(-2)^2 + 1 = -1 + 3 - 3 + 12 + 1 = (-1 + 3 - 3) + 13 = (-1) + 13 = 12. Same result. Wait, so even with exponent 1, the result is 12? That seems odd. Wait, but maybe that's just a coincidence? Wait, but in that case, the original problem would be (5x + 9) + (x + 5) + (x - 1) + 3x² + 1. Let me compute that. 5x + 9 + x + 5 + x - 1 + 3x² + 1. Combine like terms: (5x + x + x) + (9 + 5 - 1 + 1) + 3x². That is 7x + 14 + 3x². Then f(-2) would be 3*(-2)^2 + 7*(-2) + 14. Which is 3*4 -14 +14 = 12 -14 +14 = 12. So same result. So even with exponent 1, the remainder is 12. Interesting. So maybe regardless of the exponents? Wait, but in the original problem, the exponents are 611, 11, 11. But when we plug in x = -2, the first term is (-1)^611, the second term is 3^11, third term is (-3)^11. But if we had different exponents, say, even exponents, then the result would be different. But in this case, the second and third terms cancel each other because they are 3^11 and -3^11. So regardless of the exponent, as long as they are the same exponent, they would cancel? Wait, but in the original problem, the exponents are 611, 11, 11. So the second and third terms have the same exponent, 11, so they are 3^11 and (-3)^11. So they cancel. But if the exponents were different, they wouldn't. But in this case, they do. So that's why they cancel. So even if the exponent was 611, but if the second and third terms had different exponents, they wouldn't cancel. But here, they do. So that's why the second and third terms cancel. So regardless of the exponent, as long as they are the same, they would cancel. So in this case, the first term is (-1)^611, which is -1, and then the rest is 0 + 13. So total is 12. So that seems correct.

But let me check again. Maybe I made a mistake in the calculation of the first term? Let me check. 5x + 9 at x = -2: 5*(-2) is -10, plus 9 is -1. Correct. Then (-1)^611. Since 611 is odd, yes, it's -1. Correct. Then the second term: x + 5 at x = -2 is 3. Third term: x - 1 at x = -2 is -3. So 3^11 and (-3)^11. So 3^11 - 3^11 = 0. Then 3x² + 1 at x = -2: 3*(4) + 1 = 12 + 1 = 13. Then total is -1 + 0 + 13 = 12. So yes, 12. Therefore, the remainder is 12. So I think that's the answer.

But just to be thorough, let me check if there's another way to approach this problem. Maybe using modular arithmetic or something else? For example, if I consider the polynomial modulo x + 2. Since x ≡ -2 mod (x + 2), so we can substitute x = -2 into the polynomial. Which is exactly what I did. So that's the same as the Remainder Theorem. So that's consistent. So I think my approach is correct.

Alternatively, maybe I can think about expanding the terms, but that would be too tedious. For example, expanding (5x + 9)^611 would be impossible by hand. So the Remainder Theorem is the way to go here. Therefore, I think my answer is correct. So the remainder is 12. Therefore, the answer is 12. Wait, but let me check once more with another approach.

Suppose I consider the entire polynomial f(x) = (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Let me think about the terms. When divided by x + 2, the remainder is a constant, as the divisor is degree 1. So the remainder is f(-2). Which we calculated as 12. Alternatively, maybe I can check with another value? For example, if I take x = -2, and compute f(-2). But that's exactly what I did. So unless I made an arithmetic mistake in calculating 3*(-2)^2 + 1. Let me check that again. (-2)^2 is 4, times 3 is 12, plus 1 is 13. Correct. Then -1 + 13 is 12. Correct. So I think that's right.

Alternatively, maybe I can check with a different value of x. For example, suppose I take x = 0. Then f(0) = (0 + 9)^611 + (0 + 5)^11 + (0 - 1)^11 + 0 + 1. Which is 9^611 + 5^11 + (-1)^11 + 1. Which is a huge number, but if I divide by x + 2, the remainder would be f(-2) = 12. But I can't really check that here. Alternatively, maybe take x = 1. Then f(1) = (5 + 9)^611 + (1 + 5)^11 + (1 - 1)^11 + 3(1)^2 + 1. Which is 14^611 + 6^11 + 0 + 3 + 1. Again, too big to compute, but if I divide by x + 2, the remainder is still 12. But I can't verify that here. So maybe I need to trust my calculation.

Alternatively, maybe I can check if there's a mistake in the sign of the third term. Let me check again. The third term is (x - 1)^11 evaluated at x = -2. So (-2 - 1) = -3. Then (-3)^11. Since 11 is odd, that's -3^11. So yes, that's correct. So the third term is -3^11. Then the second term is 3^11. So adding them gives zero. Correct. So that's right.

Therefore, I think my answer is correct. The remainder is 12. Therefore, the answer is \boxed{12}.

**Final Answer**
\boxed{12}

To find the remainder when the polynomial
$$
f(x) = (5x + 9)^{611} + (x + 5)^{11} + (x - 1)^{11} + 3x^2 + 1
$$
is divided by $ x + 2 $, we apply the **Remainder Theorem**. This theorem states that the remainder of a polynomial $ f(x) $ divided by $ x - a $ is simply $ f(a) $. In this case, since the divisor is $ x + 2 $, we are effectively evaluating $ f(-2) $.

---

### Step 1: Evaluate $ f(-2) $

We substitute $ x = -2 $ into the polynomial:

$$
f(-2) = (5(-2) + 9)^{611} + ((-2) + 5)^{11} + ((-2) - 1)^{11} + 3(-2)^2 + 1
$$

Now compute each term:

- $ 5(-2) + 9 = -10 + 9 = -1 $, so the first term is $ (-1)^{611} = -1 $ (since 611 is odd).
- $ (-2) + 5 = 3 $, so the second term is $ 3^{11} $.
- $ (-2) - 1 = -3 $, so the third term is $ (-3)^{11} = -3^{11} $ (since 11 is odd).
- $ 3(-2)^2 = 3(4) = 12 $, and adding 1 gives $ 12 + 1 = 13 $.

---

### Step 2: Combine the terms

Now, substitute back:

$$
f(-2) = (-1) + 3^{11} + (-3^{11}) + 13
$$

Notice that $ 3^{11} - 3^{11} = 0 $, so those terms cancel out. Thus:

$$
f(-2) = -1 + 13 = 12
$$

---

### Final Answer

The remainder when the polynomial is divided by $ x + 2 $ is:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when this big polynomial is divided by x + 2. The polynomial is (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Hmm, dividing by x + 2... I remember that when you divide a polynomial by a linear divisor like x - a, the remainder is just the value of the polynomial at x = a. Wait, but here it's x + 2, which is the same as x - (-2). So, according to the Remainder Theorem, the remainder should be the value of the polynomial evaluated at x = -2. Is that right? Let me confirm. Yes, the Remainder Theorem says that if you divide a polynomial f(x) by x - a, the remainder is f(a). So in this case, since we're dividing by x + 2, which is x - (-2), the remainder is f(-2). So I just need to compute f(-2) where f(x) is that big expression. Alright, let me write that down.

So f(x) = (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Therefore, f(-2) = [5*(-2) + 9]^611 + [(-2) + 5]^11 + [(-2) - 1]^11 + 3*(-2)^2 + 1. Let me compute each term step by step.

First term: 5*(-2) + 9. Let me calculate that. 5*(-2) is -10, then adding 9 gives -10 + 9 = -1. So the first term is (-1)^611. Now, since 611 is an odd exponent, (-1)^odd is -1. So first term is -1.

Second term: (-2) + 5. That's 3. So the second term is 3^11. Let me compute 3^11. Hmm, 3^1 = 3, 3^2 = 9, 3^3 = 27, 3^4 = 81, 3^5 = 243, 3^6 = 729, 3^7 = 2187, 3^8 = 6561, 3^9 = 19683, 3^10 = 59049, 3^11 = 177147. So that's 177,147. Let me check if I can compute that another way, but maybe I can just keep it as 3^11 for now. Wait, maybe I need the exact value? Or maybe there's a simplification? Let me see.

Third term: (-2) - 1 = -3. So the third term is (-3)^11. Again, exponent is odd, so that's -3^11. Which is -177,147. Wait, same as the second term but negative. So second term is +3^11, third term is -3^11. So they might cancel each other? Let me check. If I add them together, 3^11 - 3^11 = 0? Wait, no. Wait, the second term is [(-2) + 5]^11 = 3^11, and the third term is [(-2) - 1]^11 = (-3)^11 = -3^11. So adding them together, 3^11 + (-3^11) = 0. Oh! So those two terms cancel each other out. That's nice. So the second and third terms sum to zero. That simplifies things.

So then, moving on. The first term is -1, and then the rest is 0 plus the other terms. Let me check the rest. The fourth term is 3*(-2)^2 + 1. Let me compute that. (-2)^2 is 4, so 3*4 is 12, plus 1 is 13. So altogether, f(-2) is -1 + 0 + 13. Which is 12. Therefore, the remainder is 12? Wait, let me check again.

Wait, let me retrace. First term: (-1)^611 = -1. Second term: 3^11. Third term: (-3)^11 = -3^11. So second + third term: 3^11 - 3^11 = 0. Then fourth term: 3*(-2)^2 + 1. (-2)^2 is 4, so 3*4 is 12, plus 1 is 13. So total is -1 + 0 + 13 = 12. So the remainder is 12? That seems straightforward. Wait, but let me check if I did all steps correctly.

First term: 5x + 9 at x = -2: 5*(-2) + 9 = -10 + 9 = -1. Correct. Then exponent 611: (-1)^611 = -1. Correct. Second term: x + 5 at x = -2: -2 + 5 = 3. Correct. Third term: x - 1 at x = -2: -2 -1 = -3. Correct. Then exponents: 3^11 and (-3)^11. Correct. Then 3^11 - 3^11 = 0. Then the rest: 3x² + 1. At x = -2: 3*(4) + 1 = 12 + 1 = 13. Correct. Then total: -1 + 13 = 12. So the remainder is 12? Hmm. Wait, but let me check if there's any mistake in the exponents or signs.

Wait, the original polynomial is (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. So when I plug in x = -2, it's [(-1)^611] + [3^11] + [(-3)^11] + 3*(-2)^2 + 1. Which is -1 + 3^11 - 3^11 + 12 + 1. So yes, 3^11 - 3^11 cancels, then -1 + 12 + 1 = 12. So 12. So the remainder is 12? That seems correct. Wait, but let me check if I didn't make a mistake in the exponents. For example, is the exponent on the third term 11? Yes, the original problem says (x - 1)^11. So yes. So (-3)^11 is indeed -3^11. So that's correct.

Alternatively, maybe I can check with smaller exponents? Let me try with a smaller exponent to see if my approach is correct. Suppose instead of 611, I had (5x + 9)^1. Then f(-2) would be (-1)^1 + 3^1 + (-3)^1 + 3*(-2)^2 + 1 = -1 + 3 - 3 + 12 + 1 = (-1 + 3 - 3) + 13 = (-1) + 13 = 12. Same result. Wait, so even with exponent 1, the result is 12? That seems odd. Wait, but maybe that's just a coincidence? Wait, but in that case, the original problem would be (5x + 9) + (x + 5) + (x - 1) + 3x² + 1. Let me compute that. 5x + 9 + x + 5 + x - 1 + 3x² + 1. Combine like terms: (5x + x + x) + (9 + 5 - 1 + 1) + 3x². That is 7x + 14 + 3x². Then f(-2) would be 3*(-2)^2 + 7*(-2) + 14. Which is 3*4 -14 +14 = 12 -14 +14 = 12. So same result. So even with exponent 1, the remainder is 12. Interesting. So maybe regardless of the exponents? Wait, but in the original problem, the exponents are 611, 11, 11. But when we plug in x = -2, the first term is (-1)^611, the second term is 3^11, third term is (-3)^11. But if we had different exponents, say, even exponents, then the result would be different. But in this case, the second and third terms cancel each other because they are 3^11 and -3^11. So regardless of the exponent, as long as they are the same exponent, they would cancel? Wait, but in the original problem, the exponents are 611, 11, 11. So the second and third terms have the same exponent, 11, so they are 3^11 and (-3)^11. So they cancel. But if the exponents were different, they wouldn't. But in this case, they do. So that's why they cancel. So even if the exponent was 611, but if the second and third terms had different exponents, they wouldn't cancel. But here, they do. So that's why the second and third terms cancel. So regardless of the exponent, as long as they are the same, they would cancel. So in this case, the first term is (-1)^611, which is -1, and then the rest is 0 + 13. So total is 12. So that seems correct.

But let me check again. Maybe I made a mistake in the calculation of the first term? Let me check. 5x + 9 at x = -2: 5*(-2) is -10, plus 9 is -1. Correct. Then (-1)^611. Since 611 is odd, yes, it's -1. Correct. Then the second term: x + 5 at x = -2 is 3. Third term: x - 1 at x = -2 is -3. So 3^11 and (-3)^11. So 3^11 - 3^11 = 0. Then 3x² + 1 at x = -2: 3*(4) + 1 = 12 + 1 = 13. Then total is -1 + 0 + 13 = 12. So yes, 12. Therefore, the remainder is 12. So I think that's the answer.

But just to be thorough, let me check if there's another way to approach this problem. Maybe using modular arithmetic or something else? For example, if I consider the polynomial modulo x + 2. Since x ≡ -2 mod (x + 2), so we can substitute x = -2 into the polynomial. Which is exactly what I did. So that's the same as the Remainder Theorem. So that's consistent. So I think my approach is correct.

Alternatively, maybe I can think about expanding the terms, but that would be too tedious. For example, expanding (5x + 9)^611 would be impossible by hand. So the Remainder Theorem is the way to go here. Therefore, I think my answer is correct. So the remainder is 12. Therefore, the answer is 12. Wait, but let me check once more with another approach.

Suppose I consider the entire polynomial f(x) = (5x + 9)^611 + (x + 5)^11 + (x - 1)^11 + 3x² + 1. Let me think about the terms. When divided by x + 2, the remainder is a constant, as the divisor is degree 1. So the remainder is f(-2). Which we calculated as 12. Alternatively, maybe I can check with another value? For example, if I take x = -2, and compute f(-2). But that's exactly what I did. So unless I made an arithmetic mistake in calculating 3*(-2)^2 + 1. Let me check that again. (-2)^2 is 4, times 3 is 12, plus 1 is 13. Correct. Then -1 + 13 is 12. Correct. So I think that's right.

Alternatively, maybe I can check with a different value of x. For example, suppose I take x = 0. Then f(0) = (0 + 9)^611 + (0 + 5)^11 + (0 - 1)^11 + 0 + 1. Which is 9^611 + 5^11 + (-1)^11 + 1. Which is a huge number, but if I divide by x + 2, the remainder would be f(-2) = 12. But I can't really check that here. Alternatively, maybe take x = 1. Then f(1) = (5 + 9)^611 + (1 + 5)^11 + (1 - 1)^11 + 3(1)^2 + 1. Which is 14^611 + 6^11 + 0 + 3 + 1. Again, too big to compute, but if I divide by x + 2, the remainder is still 12. But I can't verify that here. So maybe I need to trust my calculation.

Alternatively, maybe I can check if there's a mistake in the sign of the third term. Let me check again. The third term is (x - 1)^11 evaluated at x = -2. So (-2 - 1) = -3. Then (-3)^11. Since 11 is odd, that's -3^11. So yes, that's correct. So the third term is -3^11. Then the second term is 3^11. So adding them gives zero. Correct. So that's right.

Therefore, I think my answer is correct. The remainder is 12. Therefore, the answer is \boxed{12}.

**Final Answer**
\boxed{12}

To find the remainder when the polynomial
$$
f(x) = (5x + 9)^{611} + (x + 5)^{11} + (x - 1)^{11} + 3x^2 + 1
$$
is divided by $ x + 2 $, we apply the **Remainder Theorem**. This theorem states that the remainder of a polynomial $ f(x) $ divided by $ x - a $ is simply $ f(a) $. In this case, since the divisor is $ x + 2 $, we are effectively evaluating $ f(-2) $.

---

### Step 1: Evaluate $ f(-2) $

We substitute $ x = -2 $ into the polynomial:

$$
f(-2) = (5(-2) + 9)^{611} + ((-2) + 5)^{11} + ((-2) - 1)^{11} + 3(-2)^2 + 1
$$

Now compute each term:

- $ 5(-2) + 9 = -10 + 9 = -1 $, so the first term is $ (-1)^{611} = -1 $ (since 611 is odd).
- $ (-2) + 5 = 3 $, so the second term is $ 3^{11} $.
- $ (-2) - 1 = -3 $, so the third term is $ (-3)^{11} = -3^{11} $ (since 11 is odd).
- $ 3(-2)^2 = 3(4) = 12 $, and adding 1 gives $ 12 + 1 = 13 $.

---

### Step 2: Combine the terms

Now, substitute back:

$$
f(-2) = (-1) + 3^{11} + (-3^{11}) + 13
$$

Notice that $ 3^{11} - 3^{11} = 0 $, so those terms cancel out. Thus:

$$
f(-2) = -1 + 13 = 12
$$

---

### Final Answer

The remainder when the polynomial is divided by $ x + 2 $ is:

$$
\boxed{12}
$$

Solution:
By the Remainder Theorem, to find the remainder, we set $x = -2.$  This gives us
\[(-1)^{611} + 3^{11} + (-3)^{11} + 3(-2)^2 + 1 = \boxed{12}.\]


Explanation:
Okay, so I need to find the matrix A multiplied by the vector (-13, 3, 4). But I don't know what A is directly. However, I do know two things: when A is multiplied by the vector (3, 1, 0), it gives (2, 4, -3), and when multiplied by (-5, 2, 2), it gives (0, -5, -5).

Hmm, so maybe I can express the vector (-13, 3, 4) as a linear combination of the vectors (3, 1, 0) and (-5, 2, 2)? If I can do that, then I can use the linearity of matrix multiplication to find A times that vector. Let me try that approach.

Let me denote the vectors as follows:

Let’s call vector v1 = (3, 1, 0) and vector v2 = (-5, 2, 2). The target vector is v3 = (-13, 3, 4). I need to find scalars a and b such that:

a*v1 + b*v2 = v3

So, that would be:

a*(3, 1, 0) + b*(-5, 2, 2) = (-13, 3, 4)

Breaking this down into components:

For the first component: 3a - 5b = -13

Second component: 1a + 2b = 3

Third component: 0a + 2b = 4

Wait, the third component equation is 2b = 4, so that gives me b = 2. Let me check that. If I solve the third equation, 2b = 4, then b = 2. Then I can plug that into the second equation to find a.

So, second equation: a + 2b = 3. Since b = 2, then a + 4 = 3, so a = 3 - 4 = -1.

Now let me check if these values of a and b satisfy the first equation. First equation: 3a -5b = 3*(-1) -5*(2) = -3 -10 = -13. Which matches the first component of v3. Perfect! So that works.

So, the vector v3 can be written as -1*v1 + 2*v2. Therefore, applying matrix A to both sides:

A*v3 = A*(-v1 + 2*v2) = -A*v1 + 2*A*v2

We know A*v1 and A*v2 from the problem statement. Let me plug those in:

- A*v1 is - (2, 4, -3) = (-2, -4, 3)

2*A*v2 is 2*(0, -5, -5) = (0, -10, -10)

Now, adding these two results together:

(-2 + 0, -4 + (-10), 3 + (-10)) = (-2, -14, -7)

Wait, let me check that again.

First component: -2 + 0 = -2

Second component: -4 + (-10) = -14

Third component: 3 + (-10) = -7

So, the result is (-2, -14, -7). Therefore, A times the vector (-13, 3, 4) is (-2, -14, -7).

But let me verify this again step by step to make sure I didn't make any mistakes.

First, solving for a and b:

Third component equation: 2b = 4 => b = 2. Then second component: a + 2*2 = 3 => a = 3 - 4 = -1. Then first component: 3*(-1) -5*(2) = -3 -10 = -13, which matches. So that's correct.

Then, A*v3 = -A*v1 + 2*A*v2.

A*v1 is given as (2, 4, -3). So -A*v1 is (-2, -4, 3).

A*v2 is given as (0, -5, -5). So 2*A*v2 is (0, -10, -10).

Adding them together: (-2 + 0, -4 + (-10), 3 + (-10)) = (-2, -14, -7).

Yes, that seems correct.

But just to be thorough, maybe I can check if there's another way to approach this problem, or maybe check if the answer makes sense.

Alternatively, maybe I can set up the matrix A and solve for its entries? But since A is a 3x3 matrix, that would involve more variables. However, since we have two equations, maybe we can find some entries? But since we only have two vectors, maybe not enough information? Wait, but we are only asked to find A*v3, so maybe we don't need to find the entire matrix.

Alternatively, maybe I can think of the problem in terms of linear combinations. Since we have two vectors v1 and v2, and we can express v3 as a combination of them, then the result is just the combination of the images of v1 and v2 under A. Which is exactly what I did.

Alternatively, maybe I can check if the vectors v1 and v2 are linearly independent? If they are, then they form a basis for R^3? Wait, but they are in R^3, but two vectors can't form a basis for R^3. So, the vector v3 is in the span of v1 and v2, which is a plane in R^3. So, unless v3 is in that plane, which we found it is, since we could express it as a combination.

But maybe I can check if the determinant of the matrix formed by v1, v2, and some other vector is non-zero? But since we don't have a third vector, maybe not necessary.

Alternatively, maybe I can check if my answer is correct by assuming that A is a matrix with columns corresponding to the images of the standard basis vectors? Wait, but we don't have information about the standard basis vectors.

Alternatively, maybe I can think of A as a linear transformation, and since we have two vectors, we can find the transformation on any vector in the span of v1 and v2. Since v3 is in that span, we can compute it as we did.

Alternatively, maybe I can check my answer by plugging in. Suppose that A*v3 = (-2, -14, -7). Let me see if that makes sense.

Alternatively, maybe I can try to find the matrix A. Let me try that approach.

Let me denote A as a 3x3 matrix:

A = [a b c]
     [d e f]
     [g h i]

Then, multiplying A by v1 = (3, 1, 0):

A*v1 = (3a + 1b + 0c, 3d + 1e + 0f, 3g + 1h + 0i) = (3a + b, 3d + e, 3g + h) = (2, 4, -3)

Similarly, multiplying A by v2 = (-5, 2, 2):

A*v2 = (-5a + 2b + 2c, -5d + 2e + 2f, -5g + 2h + 2i) = (0, -5, -5)

So, we have the following equations:

From A*v1:

1) 3a + b = 2

2) 3d + e = 4

3) 3g + h = -3

From A*v2:

4) -5a + 2b + 2c = 0

5) -5d + 2e + 2f = -5

6) -5g + 2h + 2i = -5

So, we have 6 equations with 9 variables. So, there are infinitely many solutions. However, since we need to find A*v3, which is:

A*v3 = (-13a + 3b + 4c, -13d + 3e + 4f, -13g + 3h + 4i)

But since we don't know all the variables, maybe we can express this in terms of the equations we have.

But perhaps we can find expressions for the components of A*v3. Let me think.

Let me first try to find expressions for the variables.

From equation 1: 3a + b = 2. Let me solve for b: b = 2 - 3a.

From equation 4: -5a + 2b + 2c = 0. Substitute b from equation 1:

-5a + 2*(2 - 3a) + 2c = 0 => -5a + 4 - 6a + 2c = 0 => -11a + 4 + 2c = 0 => 2c = 11a - 4 => c = (11a - 4)/2.

Similarly, for the second column:

From equation 2: 3d + e = 4 => e = 4 - 3d.

From equation 5: -5d + 2e + 2f = -5. Substitute e:

-5d + 2*(4 - 3d) + 2f = -5 => -5d + 8 - 6d + 2f = -5 => -11d + 8 + 2f = -5 => 2f = 11d - 13 => f = (11d - 13)/2.

For the third column:

From equation 3: 3g + h = -3 => h = -3 - 3g.

From equation 6: -5g + 2h + 2i = -5. Substitute h:

-5g + 2*(-3 - 3g) + 2i = -5 => -5g -6 -6g + 2i = -5 => -11g -6 + 2i = -5 => 2i = 11g + 1 => i = (11g + 1)/2.

So, now, we can express all variables in terms of a, d, g.

Now, let's compute A*v3:

First component: -13a + 3b + 4c.

We have b = 2 - 3a, c = (11a - 4)/2.

So:

-13a + 3*(2 - 3a) + 4*( (11a - 4)/2 )

Let me compute each term:

-13a + 6 - 9a + 2*(11a - 4)

Wait, 4*( (11a - 4)/2 ) is 2*(11a - 4) = 22a - 8.

So combining:

-13a -9a + 6 + 22a - 8 = (-22a + 22a) + (6 - 8) = 0 - 2 = -2.

So the first component is -2.

Second component: -13d + 3e + 4f.

We have e = 4 - 3d, f = (11d - 13)/2.

So:

-13d + 3*(4 - 3d) + 4*( (11d - 13)/2 )

Compute each term:

-13d + 12 - 9d + 2*(11d - 13)

Which is:

-13d -9d + 12 + 22d - 26 = (-22d + 22d) + (12 - 26) = 0 -14 = -14.

Third component: -13g + 3h + 4i.

We have h = -3 - 3g, i = (11g + 1)/2.

So:

-13g + 3*(-3 - 3g) + 4*( (11g + 1)/2 )

Compute each term:

-13g -9 -9g + 2*(11g + 1)

Which is:

-13g -9g -9 + 22g + 2 = (-22g + 22g) + (-9 + 2) = 0 -7 = -7.

So, putting it all together, A*v3 = (-2, -14, -7). Which matches the result I got earlier.

Therefore, this confirms that the answer is indeed (-2, -14, -7).

So, even though I didn't find the entire matrix A, by expressing the components of A*v3 in terms of the variables and substituting the expressions derived from the given equations, I still arrived at the same result.

Therefore, I can be confident that the answer is correct.

Just to recap, the key idea was to express the target vector as a linear combination of the given vectors, then use the linearity of matrix multiplication to compute the result. This approach avoids having to determine the entire matrix A, which would be more complex and unnecessary since we only need the result for a specific vector.

Another thing to note is that even though the problem didn't specify whether the vectors v1 and v2 are linearly independent, we were able to express v3 as a combination of them, which is possible because the system of equations had a unique solution for a and b. This suggests that v1 and v2 are linearly independent? Wait, actually, since we had two vectors in R^3, they can't be linearly independent in the sense of spanning the entire space, but they can be linearly independent of each other.

Wait, let me check if v1 and v2 are linearly independent. For that, we can check if the determinant of the matrix formed by them as columns (but since they are in R^3, we need a third vector to form a square matrix). Alternatively, check if one is a scalar multiple of the other.

Looking at v1 = (3, 1, 0) and v2 = (-5, 2, 2). Suppose there exists a scalar k such that v2 = k*v1. Then:

-5 = 3k => k = -5/3

Then, the second component would be 1*k = -5/3, but the second component of v2 is 2. So, -5/3 ≠ 2. Therefore, they are not scalar multiples, so they are linearly independent.

Therefore, they form a basis for their span, which is a 2-dimensional subspace of R^3. Since v3 is in that subspace, we can express it as a combination of v1 and v2, which we did.

Therefore, this method is valid.

So, all steps check out, and the answer is consistent.

**Final Answer**
\boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}

To solve the problem, we are given two matrix-vector products:

$$
\mathbf{A} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix}, \quad \mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}
$$

We are asked to compute:

$$
\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix}
$$

---

### Step 1: Express the target vector as a linear combination

Let’s denote the vectors:

- $ \mathbf{v}_1 = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} $
- $ \mathbf{v}_2 = \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} $
- $ \mathbf{v}_3 = \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} $

We want to find scalars $ a $ and $ b $ such that:

$$
a \mathbf{v}_1 + b \mathbf{v}_2 = \mathbf{v}_3
$$

This gives the system of equations:

$$
\begin{cases}
3a - 5b = -13 \\
a + 2b = 3 \\
2b = 4
\end{cases}
$$

From the third equation: $ 2b = 4 \Rightarrow b = 2 $

Substitute into the second equation: $ a + 2(2) = 3 \Rightarrow a = -1 $

Check in the first equation: $ 3(-1) - 5(2) = -3 - 10 = -13 $, which is correct.

So, we have:

$$
\mathbf{v}_3 = -1 \cdot \mathbf{v}_1 + 2 \cdot \mathbf{v}_2
$$

---

### Step 2: Apply matrix multiplication using linearity

Since matrix multiplication is linear, we can write:

$$
\mathbf{A} \mathbf{v}_3 = \mathbf{A}(-\mathbf{v}_1 + 2\mathbf{v}_2) = -\mathbf{A} \mathbf{v}_1 + 2\mathbf{A} \mathbf{v}_2
$$

From the given:

- $ \mathbf{A} \mathbf{v}_1 = \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix} $
- $ \mathbf{A} \mathbf{v}_2 = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix} $

Now compute:

$$
-\mathbf{A} \mathbf{v}_1 = \begin{pmatrix} -2 \\ -4 \\ 3 \end{pmatrix}, \quad 2\mathbf{A} \mathbf{v}_2 = \begin{pmatrix} 0 \\ -10 \\ -10 \end{pmatrix}
$$

Add these two vectors:

$$
\begin{pmatrix} -2 \\ -4 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ -10 \\ -10 \end{pmatrix} = \begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to find the matrix A multiplied by the vector (-13, 3, 4). But I don't know what A is directly. However, I do know two things: when A is multiplied by the vector (3, 1, 0), it gives (2, 4, -3), and when multiplied by (-5, 2, 2), it gives (0, -5, -5).

Hmm, so maybe I can express the vector (-13, 3, 4) as a linear combination of the vectors (3, 1, 0) and (-5, 2, 2)? If I can do that, then I can use the linearity of matrix multiplication to find A times that vector. Let me try that approach.

Let me denote the vectors as follows:

Let’s call vector v1 = (3, 1, 0) and vector v2 = (-5, 2, 2). The target vector is v3 = (-13, 3, 4). I need to find scalars a and b such that:

a*v1 + b*v2 = v3

So, that would be:

a*(3, 1, 0) + b*(-5, 2, 2) = (-13, 3, 4)

Breaking this down into components:

For the first component: 3a - 5b = -13

Second component: 1a + 2b = 3

Third component: 0a + 2b = 4

Wait, the third component equation is 2b = 4, so that gives me b = 2. Let me check that. If I solve the third equation, 2b = 4, then b = 2. Then I can plug that into the second equation to find a.

So, second equation: a + 2b = 3. Since b = 2, then a + 4 = 3, so a = 3 - 4 = -1.

Now let me check if these values of a and b satisfy the first equation. First equation: 3a -5b = 3*(-1) -5*(2) = -3 -10 = -13. Which matches the first component of v3. Perfect! So that works.

So, the vector v3 can be written as -1*v1 + 2*v2. Therefore, applying matrix A to both sides:

A*v3 = A*(-v1 + 2*v2) = -A*v1 + 2*A*v2

We know A*v1 and A*v2 from the problem statement. Let me plug those in:

- A*v1 is - (2, 4, -3) = (-2, -4, 3)

2*A*v2 is 2*(0, -5, -5) = (0, -10, -10)

Now, adding these two results together:

(-2 + 0, -4 + (-10), 3 + (-10)) = (-2, -14, -7)

Wait, let me check that again.

First component: -2 + 0 = -2

Second component: -4 + (-10) = -14

Third component: 3 + (-10) = -7

So, the result is (-2, -14, -7). Therefore, A times the vector (-13, 3, 4) is (-2, -14, -7).

But let me verify this again step by step to make sure I didn't make any mistakes.

First, solving for a and b:

Third component equation: 2b = 4 => b = 2. Then second component: a + 2*2 = 3 => a = 3 - 4 = -1. Then first component: 3*(-1) -5*(2) = -3 -10 = -13, which matches. So that's correct.

Then, A*v3 = -A*v1 + 2*A*v2.

A*v1 is given as (2, 4, -3). So -A*v1 is (-2, -4, 3).

A*v2 is given as (0, -5, -5). So 2*A*v2 is (0, -10, -10).

Adding them together: (-2 + 0, -4 + (-10), 3 + (-10)) = (-2, -14, -7).

Yes, that seems correct.

But just to be thorough, maybe I can check if there's another way to approach this problem, or maybe check if the answer makes sense.

Alternatively, maybe I can set up the matrix A and solve for its entries? But since A is a 3x3 matrix, that would involve more variables. However, since we have two equations, maybe we can find some entries? But since we only have two vectors, maybe not enough information? Wait, but we are only asked to find A*v3, so maybe we don't need to find the entire matrix.

Alternatively, maybe I can think of the problem in terms of linear combinations. Since we have two vectors v1 and v2, and we can express v3 as a combination of them, then the result is just the combination of the images of v1 and v2 under A. Which is exactly what I did.

Alternatively, maybe I can check if the vectors v1 and v2 are linearly independent? If they are, then they form a basis for R^3? Wait, but they are in R^3, but two vectors can't form a basis for R^3. So, the vector v3 is in the span of v1 and v2, which is a plane in R^3. So, unless v3 is in that plane, which we found it is, since we could express it as a combination.

But maybe I can check if the determinant of the matrix formed by v1, v2, and some other vector is non-zero? But since we don't have a third vector, maybe not necessary.

Alternatively, maybe I can check if my answer is correct by assuming that A is a matrix with columns corresponding to the images of the standard basis vectors? Wait, but we don't have information about the standard basis vectors.

Alternatively, maybe I can think of A as a linear transformation, and since we have two vectors, we can find the transformation on any vector in the span of v1 and v2. Since v3 is in that span, we can compute it as we did.

Alternatively, maybe I can check my answer by plugging in. Suppose that A*v3 = (-2, -14, -7). Let me see if that makes sense.

Alternatively, maybe I can try to find the matrix A. Let me try that approach.

Let me denote A as a 3x3 matrix:

A = [a b c]
     [d e f]
     [g h i]

Then, multiplying A by v1 = (3, 1, 0):

A*v1 = (3a + 1b + 0c, 3d + 1e + 0f, 3g + 1h + 0i) = (3a + b, 3d + e, 3g + h) = (2, 4, -3)

Similarly, multiplying A by v2 = (-5, 2, 2):

A*v2 = (-5a + 2b + 2c, -5d + 2e + 2f, -5g + 2h + 2i) = (0, -5, -5)

So, we have the following equations:

From A*v1:

1) 3a + b = 2

2) 3d + e = 4

3) 3g + h = -3

From A*v2:

4) -5a + 2b + 2c = 0

5) -5d + 2e + 2f = -5

6) -5g + 2h + 2i = -5

So, we have 6 equations with 9 variables. So, there are infinitely many solutions. However, since we need to find A*v3, which is:

A*v3 = (-13a + 3b + 4c, -13d + 3e + 4f, -13g + 3h + 4i)

But since we don't know all the variables, maybe we can express this in terms of the equations we have.

But perhaps we can find expressions for the components of A*v3. Let me think.

Let me first try to find expressions for the variables.

From equation 1: 3a + b = 2. Let me solve for b: b = 2 - 3a.

From equation 4: -5a + 2b + 2c = 0. Substitute b from equation 1:

-5a + 2*(2 - 3a) + 2c = 0 => -5a + 4 - 6a + 2c = 0 => -11a + 4 + 2c = 0 => 2c = 11a - 4 => c = (11a - 4)/2.

Similarly, for the second column:

From equation 2: 3d + e = 4 => e = 4 - 3d.

From equation 5: -5d + 2e + 2f = -5. Substitute e:

-5d + 2*(4 - 3d) + 2f = -5 => -5d + 8 - 6d + 2f = -5 => -11d + 8 + 2f = -5 => 2f = 11d - 13 => f = (11d - 13)/2.

For the third column:

From equation 3: 3g + h = -3 => h = -3 - 3g.

From equation 6: -5g + 2h + 2i = -5. Substitute h:

-5g + 2*(-3 - 3g) + 2i = -5 => -5g -6 -6g + 2i = -5 => -11g -6 + 2i = -5 => 2i = 11g + 1 => i = (11g + 1)/2.

So, now, we can express all variables in terms of a, d, g.

Now, let's compute A*v3:

First component: -13a + 3b + 4c.

We have b = 2 - 3a, c = (11a - 4)/2.

So:

-13a + 3*(2 - 3a) + 4*( (11a - 4)/2 )

Let me compute each term:

-13a + 6 - 9a + 2*(11a - 4)

Wait, 4*( (11a - 4)/2 ) is 2*(11a - 4) = 22a - 8.

So combining:

-13a -9a + 6 + 22a - 8 = (-22a + 22a) + (6 - 8) = 0 - 2 = -2.

So the first component is -2.

Second component: -13d + 3e + 4f.

We have e = 4 - 3d, f = (11d - 13)/2.

So:

-13d + 3*(4 - 3d) + 4*( (11d - 13)/2 )

Compute each term:

-13d + 12 - 9d + 2*(11d - 13)

Which is:

-13d -9d + 12 + 22d - 26 = (-22d + 22d) + (12 - 26) = 0 -14 = -14.

Third component: -13g + 3h + 4i.

We have h = -3 - 3g, i = (11g + 1)/2.

So:

-13g + 3*(-3 - 3g) + 4*( (11g + 1)/2 )

Compute each term:

-13g -9 -9g + 2*(11g + 1)

Which is:

-13g -9g -9 + 22g + 2 = (-22g + 22g) + (-9 + 2) = 0 -7 = -7.

So, putting it all together, A*v3 = (-2, -14, -7). Which matches the result I got earlier.

Therefore, this confirms that the answer is indeed (-2, -14, -7).

So, even though I didn't find the entire matrix A, by expressing the components of A*v3 in terms of the variables and substituting the expressions derived from the given equations, I still arrived at the same result.

Therefore, I can be confident that the answer is correct.

Just to recap, the key idea was to express the target vector as a linear combination of the given vectors, then use the linearity of matrix multiplication to compute the result. This approach avoids having to determine the entire matrix A, which would be more complex and unnecessary since we only need the result for a specific vector.

Another thing to note is that even though the problem didn't specify whether the vectors v1 and v2 are linearly independent, we were able to express v3 as a combination of them, which is possible because the system of equations had a unique solution for a and b. This suggests that v1 and v2 are linearly independent? Wait, actually, since we had two vectors in R^3, they can't be linearly independent in the sense of spanning the entire space, but they can be linearly independent of each other.

Wait, let me check if v1 and v2 are linearly independent. For that, we can check if the determinant of the matrix formed by them as columns (but since they are in R^3, we need a third vector to form a square matrix). Alternatively, check if one is a scalar multiple of the other.

Looking at v1 = (3, 1, 0) and v2 = (-5, 2, 2). Suppose there exists a scalar k such that v2 = k*v1. Then:

-5 = 3k => k = -5/3

Then, the second component would be 1*k = -5/3, but the second component of v2 is 2. So, -5/3 ≠ 2. Therefore, they are not scalar multiples, so they are linearly independent.

Therefore, they form a basis for their span, which is a 2-dimensional subspace of R^3. Since v3 is in that subspace, we can express it as a combination of v1 and v2, which we did.

Therefore, this method is valid.

So, all steps check out, and the answer is consistent.

**Final Answer**
\boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}

To solve the problem, we are given two matrix-vector products:

$$
\mathbf{A} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix}, \quad \mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}
$$

We are asked to compute:

$$
\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix}
$$

---

### Step 1: Express the target vector as a linear combination

Let’s denote the vectors:

- $ \mathbf{v}_1 = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} $
- $ \mathbf{v}_2 = \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} $
- $ \mathbf{v}_3 = \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} $

We want to find scalars $ a $ and $ b $ such that:

$$
a \mathbf{v}_1 + b \mathbf{v}_2 = \mathbf{v}_3
$$

This gives the system of equations:

$$
\begin{cases}
3a - 5b = -13 \\
a + 2b = 3 \\
2b = 4
\end{cases}
$$

From the third equation: $ 2b = 4 \Rightarrow b = 2 $

Substitute into the second equation: $ a + 2(2) = 3 \Rightarrow a = -1 $

Check in the first equation: $ 3(-1) - 5(2) = -3 - 10 = -13 $, which is correct.

So, we have:

$$
\mathbf{v}_3 = -1 \cdot \mathbf{v}_1 + 2 \cdot \mathbf{v}_2
$$

---

### Step 2: Apply matrix multiplication using linearity

Since matrix multiplication is linear, we can write:

$$
\mathbf{A} \mathbf{v}_3 = \mathbf{A}(-\mathbf{v}_1 + 2\mathbf{v}_2) = -\mathbf{A} \mathbf{v}_1 + 2\mathbf{A} \mathbf{v}_2
$$

From the given:

- $ \mathbf{A} \mathbf{v}_1 = \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix} $
- $ \mathbf{A} \mathbf{v}_2 = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix} $

Now compute:

$$
-\mathbf{A} \mathbf{v}_1 = \begin{pmatrix} -2 \\ -4 \\ 3 \end{pmatrix}, \quad 2\mathbf{A} \mathbf{v}_2 = \begin{pmatrix} 0 \\ -10 \\ -10 \end{pmatrix}
$$

Add these two vectors:

$$
\begin{pmatrix} -2 \\ -4 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ -10 \\ -10 \end{pmatrix} = \begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}
$$

---

### Final Answer

$$
\boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}
$$

Solution:
Subtracting the equations $\mathbf{A} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix}$ and $\mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix},$ we get
\[\mathbf{A} \begin{pmatrix} -8 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -9 \\ -2 \end{pmatrix}.\]Then adding the equations $\mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}$ and $\mathbf{A} \begin{pmatrix} -8 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ -9 \\ -2 \end{pmatrix},$ we get
\[\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} = \boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}.\]


Explanation:
Okay, so I need to figure out how many numbers between 1 and 100 are divisible by 3, 4, and 5. Hmm, let me think. First, if a number is divisible by all three of those numbers, that means it has to be a multiple of each of them. But wait, maybe I should find the least common multiple (LCM) of 3, 4, and 5 first? Because if I can find the smallest number that is divisible by all three, then I can figure out how many multiples of that number are there up to 100.

Alright, let me recall how to find the LCM. The LCM of several numbers is the smallest number that is a multiple of each of them. To find the LCM of 3, 4, and 5, I can use their prime factorizations. Let me break each number down into prime factors:

- 3 is a prime number, so its prime factorization is just 3.
- 4 can be broken down into 2 × 2, which is 2².
- 5 is also a prime number, so its prime factorization is just 5.

To find the LCM, I need to take the highest power of each prime number that appears in the factorizations. So, looking at the primes involved: 2, 3, and 5. The highest power of 2 is 2² from 4, the highest power of 3 is 3¹ from 3, and the highest power of 5 is 5¹ from 5. Multiplying these together gives the LCM:

LCM = 2² × 3 × 5 = 4 × 3 × 5.

Let me calculate that. 4 times 3 is 12, and 12 times 5 is 60. So the LCM of 3, 4, and 5 is 60. That means any number divisible by 3, 4, and 5 must be a multiple of 60.

Now, the question is asking how many of the first 100 positive integers are divisible by 60. So I need to find how many multiples of 60 are there between 1 and 100 inclusive.

To do this, I can divide 100 by 60 and take the floor of the result. Let me do that calculation. 100 divided by 60 is equal to 1.666..., so the floor of that is 1. That would mean there is only 1 multiple of 60 within the first 100 positive integers. Wait, but let me check that.

The multiples of 60 are 60, 120, 180, etc. So within 1 to 100, the only multiple is 60 itself. 120 is already beyond 100, so yes, only 60. Therefore, the answer should be 1?

But let me make sure I didn't make a mistake in calculating the LCM. Let me verify that 60 is indeed divisible by 3, 4, and 5.

Dividing 60 by 3: 60 ÷ 3 = 20, which is an integer. Dividing 60 by 4: 60 ÷ 4 = 15, also an integer. Dividing 60 by 5: 60 ÷ 5 = 12, which is also an integer. So yes, 60 is divisible by all three numbers. Therefore, the LCM is correct.

But just to be thorough, let me think if there's another way to approach this problem. Maybe by checking numbers divisible by each of the numbers and then finding the intersection? Although that might be more complicated.

Alternatively, maybe I can use the principle of inclusion-exclusion? Wait, but inclusion-exclusion is usually for when you want to find numbers divisible by at least one of the numbers, not all of them. So maybe that's not necessary here.

Alternatively, if I think about divisibility by 3, 4, and 5, then the number must be divisible by each of them. So, as I thought before, it's equivalent to being divisible by their LCM. So that seems correct.

But let me check if there's a number less than 60 that is divisible by all three. Let's see. Let's check 30. 30 is divisible by 3 and 5, but 30 divided by 4 is 7.5, which is not an integer. How about 20? Divisible by 4 and 5, but 20 divided by 3 is about 6.666... Not an integer. How about 15? Divisible by 3 and 5, but not by 4. 12? Divisible by 3 and 4, but not by 5. 24? Divisible by 3 and 4, but not by 5. 40? Divisible by 5 and 4, but not by 3. 60 is the first one that works. So yes, 60 is indeed the smallest number divisible by all three. Therefore, the multiples are 60, 120, 180, etc. So within 1-100, only 60. Therefore, the answer is 1.

But wait, let me check if there's another multiple. Let me list all multiples of 60 up to 100. 60 × 1 = 60. 60 × 2 = 120, which is over 100. So only one multiple. Therefore, the answer is 1. So I think that's correct.

But just to make sure, let me think again. Suppose someone thought that the LCM of 3, 4, and 5 is something else. Let me recalculate the LCM. LCM of 3 and 4 is 12, right? Because 3 and 4 are co-prime? Wait, no, 3 and 4 are co-prime? Wait, their GCD is 1, so yes, LCM is 3×4=12. Then LCM of 12 and 5. Since 12 and 5 are co-prime, LCM is 12×5=60. So that's correct. So LCM is indeed 60. Therefore, the answer is 1.

Alternatively, if someone thought that LCM of 3,4,5 is 3×4×5=60, which is the same as before. So that's correct. So yeah, the answer is 1. Therefore, the number of integers between 1 and 100 divisible by 3,4,5 is 1.

But let me check once more by actually listing numbers divisible by 3, 4, and 5. Let me think. Let's take numbers divisible by 3, 4, and 5. So numbers divisible by all three. So numbers divisible by 3 and 4 and 5. So numbers divisible by 3 and 4 are numbers divisible by LCM(3,4)=12. Then numbers divisible by 12 and 5 are numbers divisible by LCM(12,5)=60. So that's the same as before. So numbers divisible by 60. So again, only 60 in 1-100. So answer is 1.

Alternatively, maybe I can check numbers divisible by 3, 4, and 5 by checking numbers divisible by 3 and 4 and 5. Let me check numbers divisible by 3 and 4 first. Numbers divisible by both 3 and 4 are numbers divisible by 12. So numbers divisible by 12 up to 100: 12, 24, 36, 48, 60, 72, 84, 96. Then from these numbers, which are also divisible by 5? So check which of these numbers are divisible by 5. Let's see:

12: 12 ÷ 5 = 2.4 → nope.

24 ÷ 5 = 4.8 → nope.

36 ÷ 5 = 7.2 → nope.

48 ÷ 5 = 9.6 → nope.

60 ÷ 5 = 12 → yes.

72 ÷ 5 = 14.4 → nope.

84 ÷ 5 = 16.8 → nope.

96 ÷ 5 = 19.2 → nope.

So only 60 is divisible by 5. Therefore, only one number. Therefore, the answer is 1. So that confirms it again.

Alternatively, maybe I can think of numbers divisible by 5 and 4 first. Numbers divisible by 5 and 4 are numbers divisible by LCM(5,4)=20. So numbers divisible by 20 up to 100: 20, 40, 60, 80, 100. Then check which of these are divisible by 3. Let's check:

20 ÷ 3 ≈ 6.666... nope.

40 ÷ 3 ≈ 13.333... nope.

60 ÷ 3 = 20 → yes.

80 ÷ 3 ≈ 26.666... nope.

100 ÷ 3 ≈ 33.333... nope.

So again, only 60. Therefore, answer is 1. So that's another way to check.

Alternatively, numbers divisible by 3 and 5 are numbers divisible by 15. Then check which of those are divisible by 4. Let's see. Numbers divisible by 15 up to 100: 15, 30, 45, 60, 75, 90. Check which are divisible by 4:

15 ÷ 4 = 3.75 → nope.

30 ÷ 4 = 7.5 → nope.

45 ÷ 4 = 11.25 → nope.

60 ÷ 4 = 15 → yes.

75 ÷ 4 = 18.75 → nope.

90 ÷ 4 = 22.5 → nope.

Again, only 60. So same result. Therefore, regardless of the method, the answer is 1. Therefore, I can be confident that the answer is 1.

But just to make sure, let me think if there's any other number between 1 and 100 that is divisible by 3, 4, and 5. Let me think of 60. 60 is 3×4×5, but is there another number? Let me think of 60×2=120, which is over 100. So no. So only 60. Therefore, the answer is 1.

Alternatively, maybe I can think of the formula for the number of multiples of a number n up to some limit. The formula is floor(limit / n). So here, the limit is 100, and n is 60. So floor(100 / 60) = floor(1.666...) = 1. So that's another way to get the answer. So that's consistent.

Therefore, after checking multiple ways, I think the answer is 1. Therefore, the number of integers from 1 to 100 divisible by 3, 4, and 5 is 1.

**Final Answer**
The number of integers is \boxed{1}.

To determine how many of the first one hundred positive integers are divisible by **3, 4, and 5**, we need to find the numbers that are divisible by **all three** of these numbers simultaneously.

---

### Step 1: Understand the Problem

A number divisible by **3, 4, and 5** must be divisible by **all three**. This means it must be a **common multiple** of 3, 4, and 5. The **least common multiple (LCM)** of these numbers is the smallest such number, and all other numbers divisible by all three will be multiples of this LCM.

---

### Step 2: Find the LCM of 3, 4, and 5

We can find the LCM by using the **prime factorization** method:

- $ 3 = 3 $
- $ 4 = 2^2 $
- $ 5 = 5 $

The LCM is the product of the **highest powers** of all prime factors involved:

$$
\text{LCM}(3, 4, 5) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60
$$

So, any number divisible by 3, 4, and 5 must be a **multiple of 60**.

---

### Step 3: Count the Multiples of 60 Between 1 and 100

We now count how many **multiples of 60** are less than or equal to 100.

$$
\left\lfloor \frac{100}{60} \right\rfloor = \left\lfloor 1.666\ldots \right\rfloor = 1
$$

This means there is **only one** multiple of 60 within the range from 1 to 100, and that is:

$$
60 \times 1 = 60
$$

---

### Step 4: Verify the Result

Let’s confirm that 60 is indeed divisible by 3, 4, and 5:

- $ 60 \div 3 = 20 $ → integer
- $ 60 \div 4 = 15 $ → integer
- $ 60 \div 5 = 12 $ → integer

So, 60 is divisible by all three numbers. The next multiple, $ 60 \times 2 = 120 $, is greater than 100, so it is not included.

---

### Final Answer

$$
\boxed{1}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many numbers between 1 and 100 are divisible by 3, 4, and 5. Hmm, let me think. First, if a number is divisible by all three of those numbers, that means it has to be a multiple of each of them. But wait, maybe I should find the least common multiple (LCM) of 3, 4, and 5 first? Because if I can find the smallest number that is divisible by all three, then I can figure out how many multiples of that number are there up to 100.

Alright, let me recall how to find the LCM. The LCM of several numbers is the smallest number that is a multiple of each of them. To find the LCM of 3, 4, and 5, I can use their prime factorizations. Let me break each number down into prime factors:

- 3 is a prime number, so its prime factorization is just 3.
- 4 can be broken down into 2 × 2, which is 2².
- 5 is also a prime number, so its prime factorization is just 5.

To find the LCM, I need to take the highest power of each prime number that appears in the factorizations. So, looking at the primes involved: 2, 3, and 5. The highest power of 2 is 2² from 4, the highest power of 3 is 3¹ from 3, and the highest power of 5 is 5¹ from 5. Multiplying these together gives the LCM:

LCM = 2² × 3 × 5 = 4 × 3 × 5.

Let me calculate that. 4 times 3 is 12, and 12 times 5 is 60. So the LCM of 3, 4, and 5 is 60. That means any number divisible by 3, 4, and 5 must be a multiple of 60.

Now, the question is asking how many of the first 100 positive integers are divisible by 60. So I need to find how many multiples of 60 are there between 1 and 100 inclusive.

To do this, I can divide 100 by 60 and take the floor of the result. Let me do that calculation. 100 divided by 60 is equal to 1.666..., so the floor of that is 1. That would mean there is only 1 multiple of 60 within the first 100 positive integers. Wait, but let me check that.

The multiples of 60 are 60, 120, 180, etc. So within 1 to 100, the only multiple is 60 itself. 120 is already beyond 100, so yes, only 60. Therefore, the answer should be 1?

But let me make sure I didn't make a mistake in calculating the LCM. Let me verify that 60 is indeed divisible by 3, 4, and 5.

Dividing 60 by 3: 60 ÷ 3 = 20, which is an integer. Dividing 60 by 4: 60 ÷ 4 = 15, also an integer. Dividing 60 by 5: 60 ÷ 5 = 12, which is also an integer. So yes, 60 is divisible by all three numbers. Therefore, the LCM is correct.

But just to be thorough, let me think if there's another way to approach this problem. Maybe by checking numbers divisible by each of the numbers and then finding the intersection? Although that might be more complicated.

Alternatively, maybe I can use the principle of inclusion-exclusion? Wait, but inclusion-exclusion is usually for when you want to find numbers divisible by at least one of the numbers, not all of them. So maybe that's not necessary here.

Alternatively, if I think about divisibility by 3, 4, and 5, then the number must be divisible by each of them. So, as I thought before, it's equivalent to being divisible by their LCM. So that seems correct.

But let me check if there's a number less than 60 that is divisible by all three. Let's see. Let's check 30. 30 is divisible by 3 and 5, but 30 divided by 4 is 7.5, which is not an integer. How about 20? Divisible by 4 and 5, but 20 divided by 3 is about 6.666... Not an integer. How about 15? Divisible by 3 and 5, but not by 4. 12? Divisible by 3 and 4, but not by 5. 24? Divisible by 3 and 4, but not by 5. 40? Divisible by 5 and 4, but not by 3. 60 is the first one that works. So yes, 60 is indeed the smallest number divisible by all three. Therefore, the multiples are 60, 120, 180, etc. So within 1-100, only 60. Therefore, the answer is 1.

But wait, let me check if there's another multiple. Let me list all multiples of 60 up to 100. 60 × 1 = 60. 60 × 2 = 120, which is over 100. So only one multiple. Therefore, the answer is 1. So I think that's correct.

But just to make sure, let me think again. Suppose someone thought that the LCM of 3, 4, and 5 is something else. Let me recalculate the LCM. LCM of 3 and 4 is 12, right? Because 3 and 4 are co-prime? Wait, no, 3 and 4 are co-prime? Wait, their GCD is 1, so yes, LCM is 3×4=12. Then LCM of 12 and 5. Since 12 and 5 are co-prime, LCM is 12×5=60. So that's correct. So LCM is indeed 60. Therefore, the answer is 1.

Alternatively, if someone thought that LCM of 3,4,5 is 3×4×5=60, which is the same as before. So that's correct. So yeah, the answer is 1. Therefore, the number of integers between 1 and 100 divisible by 3,4,5 is 1.

But let me check once more by actually listing numbers divisible by 3, 4, and 5. Let me think. Let's take numbers divisible by 3, 4, and 5. So numbers divisible by all three. So numbers divisible by 3 and 4 and 5. So numbers divisible by 3 and 4 are numbers divisible by LCM(3,4)=12. Then numbers divisible by 12 and 5 are numbers divisible by LCM(12,5)=60. So that's the same as before. So numbers divisible by 60. So again, only 60 in 1-100. So answer is 1.

Alternatively, maybe I can check numbers divisible by 3, 4, and 5 by checking numbers divisible by 3 and 4 and 5. Let me check numbers divisible by 3 and 4 first. Numbers divisible by both 3 and 4 are numbers divisible by 12. So numbers divisible by 12 up to 100: 12, 24, 36, 48, 60, 72, 84, 96. Then from these numbers, which are also divisible by 5? So check which of these numbers are divisible by 5. Let's see:

12: 12 ÷ 5 = 2.4 → nope.

24 ÷ 5 = 4.8 → nope.

36 ÷ 5 = 7.2 → nope.

48 ÷ 5 = 9.6 → nope.

60 ÷ 5 = 12 → yes.

72 ÷ 5 = 14.4 → nope.

84 ÷ 5 = 16.8 → nope.

96 ÷ 5 = 19.2 → nope.

So only 60 is divisible by 5. Therefore, only one number. Therefore, the answer is 1. So that confirms it again.

Alternatively, maybe I can think of numbers divisible by 5 and 4 first. Numbers divisible by 5 and 4 are numbers divisible by LCM(5,4)=20. So numbers divisible by 20 up to 100: 20, 40, 60, 80, 100. Then check which of these are divisible by 3. Let's check:

20 ÷ 3 ≈ 6.666... nope.

40 ÷ 3 ≈ 13.333... nope.

60 ÷ 3 = 20 → yes.

80 ÷ 3 ≈ 26.666... nope.

100 ÷ 3 ≈ 33.333... nope.

So again, only 60. Therefore, answer is 1. So that's another way to check.

Alternatively, numbers divisible by 3 and 5 are numbers divisible by 15. Then check which of those are divisible by 4. Let's see. Numbers divisible by 15 up to 100: 15, 30, 45, 60, 75, 90. Check which are divisible by 4:

15 ÷ 4 = 3.75 → nope.

30 ÷ 4 = 7.5 → nope.

45 ÷ 4 = 11.25 → nope.

60 ÷ 4 = 15 → yes.

75 ÷ 4 = 18.75 → nope.

90 ÷ 4 = 22.5 → nope.

Again, only 60. So same result. Therefore, regardless of the method, the answer is 1. Therefore, I can be confident that the answer is 1.

But just to make sure, let me think if there's any other number between 1 and 100 that is divisible by 3, 4, and 5. Let me think of 60. 60 is 3×4×5, but is there another number? Let me think of 60×2=120, which is over 100. So no. So only 60. Therefore, the answer is 1.

Alternatively, maybe I can think of the formula for the number of multiples of a number n up to some limit. The formula is floor(limit / n). So here, the limit is 100, and n is 60. So floor(100 / 60) = floor(1.666...) = 1. So that's another way to get the answer. So that's consistent.

Therefore, after checking multiple ways, I think the answer is 1. Therefore, the number of integers from 1 to 100 divisible by 3, 4, and 5 is 1.

**Final Answer**
The number of integers is \boxed{1}.

To determine how many of the first one hundred positive integers are divisible by **3, 4, and 5**, we need to find the numbers that are divisible by **all three** of these numbers simultaneously.

---

### Step 1: Understand the Problem

A number divisible by **3, 4, and 5** must be divisible by **all three**. This means it must be a **common multiple** of 3, 4, and 5. The **least common multiple (LCM)** of these numbers is the smallest such number, and all other numbers divisible by all three will be multiples of this LCM.

---

### Step 2: Find the LCM of 3, 4, and 5

We can find the LCM by using the **prime factorization** method:

- $ 3 = 3 $
- $ 4 = 2^2 $
- $ 5 = 5 $

The LCM is the product of the **highest powers** of all prime factors involved:

$$
\text{LCM}(3, 4, 5) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60
$$

So, any number divisible by 3, 4, and 5 must be a **multiple of 60**.

---

### Step 3: Count the Multiples of 60 Between 1 and 100

We now count how many **multiples of 60** are less than or equal to 100.

$$
\left\lfloor \frac{100}{60} \right\rfloor = \left\lfloor 1.666\ldots \right\rfloor = 1
$$

This means there is **only one** multiple of 60 within the range from 1 to 100, and that is:

$$
60 \times 1 = 60
$$

---

### Step 4: Verify the Result

Let’s confirm that 60 is indeed divisible by 3, 4, and 5:

- $ 60 \div 3 = 20 $ → integer
- $ 60 \div 4 = 15 $ → integer
- $ 60 \div 5 = 12 $ → integer

So, 60 is divisible by all three numbers. The next multiple, $ 60 \times 2 = 120 $, is greater than 100, so it is not included.

---

### Final Answer

$$
\boxed{1}
$$

Solution: We could do this by the divisibility rules, but that would be quite tedious. It's easier to note that a number divisible by $3, 4,$ and $5$ must be divisible by their product, $3 \times 4 \times 5 = 60$. This is because a number which is divisible by several integers must be divisible by their least common multiple -- however, since $3, 4,$ and $5$ are relatively prime, the least common multiple is just the product of all three. Clearly, there is only one number between $1$ and $100$ divisible by $60;$ that is, $60$ itself. Thus there is only $\boxed{1}$ such number.


Explanation:
Okay, let's see. Frank took a 100-question test. Each correct answer gives 5 points, each incorrect answer subtracts 2 points, and unanswered questions give 0. He answered 80 questions and scored 232 points. I need to find out how many he answered correctly.

First, let me parse the problem again. Total questions: 100. He answered 80, so that means he left 20 questions unanswered. But the scoring is based on correct, incorrect, and unanswered. Since unanswered questions give zero, maybe I don't need to worry about them? But maybe they are part of the total? Wait, the problem says he answered 80 questions. So, the remaining 20 were not answered.

So, the total number of questions he attempted is 80. Out of these 80, some are correct and some are incorrect. Let me denote the number of correct answers as, let's say, 'c'. Then the number of incorrect answers would be total answered minus correct, so 80 - c.

So, his total score would be 5 times correct answers minus 2 times incorrect answers. And that equals 232.

So, the equation would be: 5c - 2(80 - c) = 232.

Let me write that down:

5c - 2(80 - c) = 232

Let me solve this equation step by step.

First, expand the terms:

5c - 160 + 2c = 232. Wait, because -2*(80 - c) is -160 + 2c.

So combining like terms: 5c + 2c is 7c. So:

7c - 160 = 232

Then, adding 160 to both sides:

7c = 232 + 160

Let me calculate 232 + 160. 200 + 100 is 300, 32 + 60 is 92, so total is 392.

So 7c = 392. Therefore, c = 392 / 7. Let me compute that.

Dividing 392 by 7. 7*50 is 350. 392 - 350 is 42. 42 /7 is 6. So total is 50 + 6 = 56.

So c = 56.

Wait, so Frank answered 56 correctly? Let me check if that makes sense.

If he answered 56 correctly, then incorrect answers are 80 - 56 = 24.

So his score would be 5*56 - 2*24. Let me compute that.

5*56 is 280. 2*24 is 48. So 280 - 48 = 232. Which matches the given score.

So that seems correct.

But let me think again if I considered all the information properly. The total number of questions is 100, but he answered 80. So 20 were left blank. But since blank questions give zero, they don't affect the score. So the score is only based on the 80 answered questions.

Therefore, my equation is correct.

Alternatively, maybe I can think of it as variables. Let me try another approach to verify.

Let me denote:

Let c = correct answers

i = incorrect answers

u = unanswered questions

We know that c + i + u = 100.

But Frank answered 80 questions, so c + i = 80. Therefore, u = 100 - 80 = 20. Which matches the earlier thought.

His score is 5c - 2i + 0*u = 232.

So, 5c - 2i = 232.

But since c + i = 80, we can write i = 80 - c.

Substituting into the score equation:

5c - 2(80 - c) = 232. Which is exactly the same equation as before.

So solving gives c = 56.

Therefore, the answer is 56.

But let me check again.

If he got 56 correct, 24 incorrect, then total score is 5*56 = 280, minus 2*24 = 48, so 280 - 48 = 232. Correct.

Alternatively, if I thought maybe there's another way to approach it? Maybe using some other method?

Alternatively, think about the total points if all answered questions were correct. If all 80 answered questions were correct, he would have 80*5 = 400 points. But he has 232 points. So the difference is 400 - 232 = 168.

But each incorrect answer is not only not giving 5 points but also subtracting 2. So, compared to a correct answer, an incorrect answer is worth 5 - (-2) = 7 points less? Wait, no. Wait, if you have an incorrect answer, instead of gaining 5, you lose 2. So the difference between a correct and incorrect is 5 - (-2) = 7? Wait, maybe not. Let me think.

Suppose instead of having a correct answer, you have an incorrect one. So, the difference in points would be ( -2 ) - 5 = -7. So, each incorrect answer is 7 points less than a correct answer.

But how does that help?

Let me think. Let me assume that all 80 answers were correct, which would be 400 points. But he has 232. So, the difference is 400 - 232 = 168.

Each incorrect answer compared to a correct one reduces the total by 7 points. So, if there are x incorrect answers, then total reduction is 7x. Therefore, 7x = 168. Therefore, x = 24. Therefore, incorrect answers are 24, correct answers are 80 - 24 = 56.

Same answer. So that's another way to think about it.

Therefore, seems like 56 is correct.

Alternatively, maybe using algebra again.

Let me try to set up equations again.

Let me define variables:

Let c = number of correct answers

Let i = number of incorrect answers

We know that c + i = 80

And 5c - 2i = 232

So solving these equations.

From first equation: i = 80 - c. Substitute into second equation:

5c - 2(80 - c) = 232

Which is same as before.

So, 5c - 160 + 2c = 232

7c = 392

c = 56.

Same result.

Therefore, seems like 56 is correct.

Alternatively, maybe think in terms of systems of equations.

But I think that's redundant.

Alternatively, think about the total points.

But I think all approaches lead to 56.

Therefore, I think the answer is 56.

But let me check once more.

Total correct: 56, total incorrect: 24.

Total points: 56*5 = 280.

24*2 = 48.

280 - 48 = 232.

Yes.

So, that's correct.

Therefore, Frank answered 56 questions correctly.

**Final Answer**
The number of questions Frank answered correctly is \boxed{56}.

To determine how many questions Frank answered correctly, we begin by analyzing the structure of the test and the scoring system:

- **Total questions**: 100
- **Correct answer**: +5 points
- **Incorrect answer**: -2 points
- **Unanswered question**: 0 points
- **Frank answered**: 80 questions
- **Frank's total score**: 232 points

---

### Step 1: Define Variables

Let:
- $ c $ = number of correct answers
- $ i $ = number of incorrect answers

From the problem, we know:
- $ c + i = 80 $ (since Frank answered 80 questions)
- $ 5c - 2i = 232 $ (based on the scoring system)

---

### Step 2: Solve the System of Equations

From the first equation:
$$
i = 80 - c
$$

Substitute into the second equation:
$$
5c - 2(80 - c) = 232
$$

Simplify:
$$
5c - 160 + 2c = 232
$$
$$
7c - 160 = 232
$$
$$
7c = 392
$$
$$
c = \frac{392}{7} = 56
$$

---

### Step 3: Verify the Solution

If Frank answered 56 questions correctly:
- Incorrect answers = $ 80 - 56 = 24 $
- Score = $ 5 \times 56 - 2 \times 24 = 280 - 48 = 232 $

This matches the given total score, confirming the solution is correct.

---

### Final Answer

$$
\boxed{56}
$$

[Assistant]
Explanation:
Okay, let's see. Frank took a 100-question test. Each correct answer gives 5 points, each incorrect answer subtracts 2 points, and unanswered questions give 0. He answered 80 questions and scored 232 points. I need to find out how many he answered correctly.

First, let me parse the problem again. Total questions: 100. He answered 80, so that means he left 20 questions unanswered. But the scoring is based on correct, incorrect, and unanswered. Since unanswered questions give zero, maybe I don't need to worry about them? But maybe they are part of the total? Wait, the problem says he answered 80 questions. So, the remaining 20 were not answered.

So, the total number of questions he attempted is 80. Out of these 80, some are correct and some are incorrect. Let me denote the number of correct answers as, let's say, 'c'. Then the number of incorrect answers would be total answered minus correct, so 80 - c.

So, his total score would be 5 times correct answers minus 2 times incorrect answers. And that equals 232.

So, the equation would be: 5c - 2(80 - c) = 232.

Let me write that down:

5c - 2(80 - c) = 232

Let me solve this equation step by step.

First, expand the terms:

5c - 160 + 2c = 232. Wait, because -2*(80 - c) is -160 + 2c.

So combining like terms: 5c + 2c is 7c. So:

7c - 160 = 232

Then, adding 160 to both sides:

7c = 232 + 160

Let me calculate 232 + 160. 200 + 100 is 300, 32 + 60 is 92, so total is 392.

So 7c = 392. Therefore, c = 392 / 7. Let me compute that.

Dividing 392 by 7. 7*50 is 350. 392 - 350 is 42. 42 /7 is 6. So total is 50 + 6 = 56.

So c = 56.

Wait, so Frank answered 56 correctly? Let me check if that makes sense.

If he answered 56 correctly, then incorrect answers are 80 - 56 = 24.

So his score would be 5*56 - 2*24. Let me compute that.

5*56 is 280. 2*24 is 48. So 280 - 48 = 232. Which matches the given score.

So that seems correct.

But let me think again if I considered all the information properly. The total number of questions is 100, but he answered 80. So 20 were left blank. But since blank questions give zero, they don't affect the score. So the score is only based on the 80 answered questions.

Therefore, my equation is correct.

Alternatively, maybe I can think of it as variables. Let me try another approach to verify.

Let me denote:

Let c = correct answers

i = incorrect answers

u = unanswered questions

We know that c + i + u = 100.

But Frank answered 80 questions, so c + i = 80. Therefore, u = 100 - 80 = 20. Which matches the earlier thought.

His score is 5c - 2i + 0*u = 232.

So, 5c - 2i = 232.

But since c + i = 80, we can write i = 80 - c.

Substituting into the score equation:

5c - 2(80 - c) = 232. Which is exactly the same equation as before.

So solving gives c = 56.

Therefore, the answer is 56.

But let me check again.

If he got 56 correct, 24 incorrect, then total score is 5*56 = 280, minus 2*24 = 48, so 280 - 48 = 232. Correct.

Alternatively, if I thought maybe there's another way to approach it? Maybe using some other method?

Alternatively, think about the total points if all answered questions were correct. If all 80 answered questions were correct, he would have 80*5 = 400 points. But he has 232 points. So the difference is 400 - 232 = 168.

But each incorrect answer is not only not giving 5 points but also subtracting 2. So, compared to a correct answer, an incorrect answer is worth 5 - (-2) = 7 points less? Wait, no. Wait, if you have an incorrect answer, instead of gaining 5, you lose 2. So the difference between a correct and incorrect is 5 - (-2) = 7? Wait, maybe not. Let me think.

Suppose instead of having a correct answer, you have an incorrect one. So, the difference in points would be ( -2 ) - 5 = -7. So, each incorrect answer is 7 points less than a correct answer.

But how does that help?

Let me think. Let me assume that all 80 answers were correct, which would be 400 points. But he has 232. So, the difference is 400 - 232 = 168.

Each incorrect answer compared to a correct one reduces the total by 7 points. So, if there are x incorrect answers, then total reduction is 7x. Therefore, 7x = 168. Therefore, x = 24. Therefore, incorrect answers are 24, correct answers are 80 - 24 = 56.

Same answer. So that's another way to think about it.

Therefore, seems like 56 is correct.

Alternatively, maybe using algebra again.

Let me try to set up equations again.

Let me define variables:

Let c = number of correct answers

Let i = number of incorrect answers

We know that c + i = 80

And 5c - 2i = 232

So solving these equations.

From first equation: i = 80 - c. Substitute into second equation:

5c - 2(80 - c) = 232

Which is same as before.

So, 5c - 160 + 2c = 232

7c = 392

c = 56.

Same result.

Therefore, seems like 56 is correct.

Alternatively, maybe think in terms of systems of equations.

But I think that's redundant.

Alternatively, think about the total points.

But I think all approaches lead to 56.

Therefore, I think the answer is 56.

But let me check once more.

Total correct: 56, total incorrect: 24.

Total points: 56*5 = 280.

24*2 = 48.

280 - 48 = 232.

Yes.

So, that's correct.

Therefore, Frank answered 56 questions correctly.

**Final Answer**
The number of questions Frank answered correctly is \boxed{56}.

To determine how many questions Frank answered correctly, we begin by analyzing the structure of the test and the scoring system:

- **Total questions**: 100
- **Correct answer**: +5 points
- **Incorrect answer**: -2 points
- **Unanswered question**: 0 points
- **Frank answered**: 80 questions
- **Frank's total score**: 232 points

---

### Step 1: Define Variables

Let:
- $ c $ = number of correct answers
- $ i $ = number of incorrect answers

From the problem, we know:
- $ c + i = 80 $ (since Frank answered 80 questions)
- $ 5c - 2i = 232 $ (based on the scoring system)

---

### Step 2: Solve the System of Equations

From the first equation:
$$
i = 80 - c
$$

Substitute into the second equation:
$$
5c - 2(80 - c) = 232
$$

Simplify:
$$
5c - 160 + 2c = 232
$$
$$
7c - 160 = 232
$$
$$
7c = 392
$$
$$
c = \frac{392}{7} = 56
$$

---

### Step 3: Verify the Solution

If Frank answered 56 questions correctly:
- Incorrect answers = $ 80 - 56 = 24 $
- Score = $ 5 \times 56 - 2 \times 24 = 280 - 48 = 232 $

This matches the given total score, confirming the solution is correct.

---

### Final Answer

$$
\boxed{56}
$$

Solution:
Let the number of questions Frank answered correctly be $a$ and the number of questions he answered incorrectly be $b$. We have the two equations \begin{align*}
a+b&=80\\
5a-2b&=232
\end{align*} From the first equation, we have $b=80-a$. Substituting this into the second equation to eliminate $b$, we have $5a-2(80-a)=232\Rightarrow a=56$. Thus, Frank answered $\boxed{56}$ questions correctly.