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 matrix A is. 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 figure out the matrix A by using these two equations? But wait, matrix A is a 3x3 matrix, right? Because the vectors are in 3-dimensional space. So if I can express the vector (-13, 3, 4) as a linear combination of the vectors (3, 1, 0) and (-5, 2, 2), then I can use the linearity of matrix multiplication to find A times that vector.

But wait, hold on. The problem is that we have two vectors, but we need to express a third vector as a combination of them. However, since we have two vectors in 3D space, they might not span the entire space. So maybe there's another vector we need to consider? Or maybe there's a way to set up equations?

Alternatively, maybe I can set up a system of equations. Let me think. Let me denote the matrix A as:

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

But that's a lot of variables. However, if I can use the given information, maybe I can find some relations between the entries of A.

First, let's consider the first equation: A * [3, 1, 0]^T = [2, 4, -3]^T.

So, multiplying A by the vector (3, 1, 0) gives:

First component: 3a + 1b + 0c = 2
Second component: 3d + 1e + 0f = 4
Third component: 3g + 1h + 0i = -3

So that gives us three equations:

1) 3a + b = 2
2) 3d + e = 4
3) 3g + h = -3

Similarly, the second equation: A * [-5, 2, 2]^T = [0, -5, -5]^T.

So multiplying A by (-5, 2, 2):

First component: -5a + 2b + 2c = 0
Second component: -5d + 2e + 2f = -5
Third component: -5g + 2h + 2i = -5

So that gives us three more equations:

4) -5a + 2b + 2c = 0
5) -5d + 2e + 2f = -5
6) -5g + 2h + 2i = -5

So now, we have six equations, but we have nine variables (a, b, c, d, e, f, g, h, i). So we need more information. But the problem is asking for A times another vector, which is (-13, 3, 4). Maybe we can express (-13, 3, 4) as a linear combination of the two given vectors, and then use linearity?

Let me try that approach. Let me denote the first vector as v1 = [3, 1, 0]^T and the second vector as v2 = [-5, 2, 2]^T. Let me see if I can find scalars x and y such that:

x*v1 + y*v2 = [-13, 3, 4]^T

So, writing that out:

For the first component: 3x -5y = -13
Second component: 1x + 2y = 3
Third component: 0x + 2y = 4

So, let's solve this system. Let me start with the third component equation: 2y = 4 => y = 2.

Then, substitute y = 2 into the second component equation: x + 2*2 = 3 => x + 4 = 3 => x = 3 - 4 = -1.

Now check if these values satisfy the first component equation: 3*(-1) -5*(2) = -3 -10 = -13. Yes! Perfect. So x = -1 and y = 2.

Therefore, the vector [-13, 3, 4]^T is equal to -1*v1 + 2*v2.

Therefore, by linearity of matrix multiplication, A*[-13, 3, 4]^T = -1*A*v1 + 2*A*v2.

We know A*v1 and A*v2 from the problem statement.

So, substituting:

-1*[2, 4, -3]^T + 2*[0, -5, -5]^T

Let me compute each component:

First component: -1*2 + 2*0 = -2 + 0 = -2

Second component: -1*4 + 2*(-5) = -4 -10 = -14

Third component: -1*(-3) + 2*(-5) = 3 -10 = -7

Therefore, the result is [-2, -14, -7]^T.

Wait, but let me check again. Let me recompute:

First component: -1*(2) + 2*(0) = -2 + 0 = -2

Second component: -1*(4) + 2*(-5) = -4 -10 = -14

Third component: -1*(-3) + 2*(-5) = 3 -10 = -7

Yes, that seems correct. So the answer should be the vector (-2, -14, -7).

But let me make sure that I didn't make any mistakes in solving for x and y. Let me check again:

Original equations:

3x -5y = -13

x + 2y = 3

2y = 4 => y = 2. Then x = 3 - 2y = 3 - 4 = -1. Then 3x -5y = 3*(-1) -5*(2) = -3 -10 = -13. Correct.

So that's correct. Therefore, the linear combination is correct.

Therefore, the answer is [-2, -14, -7]^T.

But let me think again. Is there another way to check this? Maybe by trying to find the matrix A? Although that would require more work. Let me try to see if I can find some entries of A.

From the first set of equations:

From equation 1: 3a + b = 2

From equation 4: -5a + 2b + 2c = 0

Similarly, for the second vector, equations 2 and 5:

Equation 2: 3d + e = 4

Equation 5: -5d + 2e + 2f = -5

And for the third component:

Equation 3: 3g + h = -3

Equation 6: -5g + 2h + 2i = -5

So, for the first column, we have equations 1 and 4. Let me try to solve for a, b, c.

From equation 1: 3a + b = 2 => b = 2 - 3a

Substitute into equation 4: -5a + 2*(2 - 3a) + 2c = 0 => -5a + 4 -6a + 2c = 0 => -11a + 4 + 2c = 0 => 2c = 11a -4 => c = (11a -4)/2

But we have three variables here (a, b, c) but only two equations. So we can't determine them uniquely. Similarly for the other columns.

Therefore, the matrix A is not uniquely determined by the given information. However, since we are only asked to compute A times a specific vector, which can be expressed as a linear combination of the given vectors, we don't need to find the entire matrix.

Therefore, our previous approach is valid.

But just to be thorough, let me check if there's any inconsistency. Suppose I pick a value for a, then compute b and c. Let me pick a value for a. Let me choose a = 0. Then b = 2 - 0 = 2. Then c = (0 -4)/2 = -2.

So then, if a = 0, b = 2, c = -2.

Then, let me check if this works with equation 4: -5*0 + 2*2 + 2*(-2) = 0 + 4 -4 = 0. Correct.

So that works.

Similarly, if I pick another value for a, say a = 1. Then b = 2 - 3*1 = -1. Then c = (11*1 -4)/2 = (7)/2 = 3.5.

So then, equation 4: -5*1 + 2*(-1) + 2*(3.5) = -5 -2 +7 = 0. Correct.

So, there are infinitely many solutions for the first column. However, when we apply A to the vector [-13, 3, 4], which is a combination of v1 and v2, the result only depends on the action of A on v1 and v2, which are given. Therefore, regardless of the other entries of A, the result is determined.

Therefore, our answer is correct.

Alternatively, if I wanted to find the entire matrix A, I would need more information. But since we don't have it, and the problem only requires a specific application of A, our method is valid.

Therefore, the answer is the vector (-2, -14, -7).

But let me check once more by another approach. Suppose I try to find A. Let me think.

Since we have two vectors, maybe we can find A by setting up equations. However, since we have two vectors, we can't determine A uniquely. However, if we assume that the third column is arbitrary? Wait, but the problem is that the vector [-13, 3, 4] is in the span of v1 and v2, so even if we don't know the third column, since we are not using it in the calculation, it's okay.

Alternatively, maybe I can think of A as a linear transformation, and since we know its action on two vectors, we can find its action on any vector in their span. Since [-13, 3, 4] is in their span, we can compute it as we did.

Therefore, I think the answer is correct.

Wait, but let me check once again the calculation of -1*(2,4,-3) + 2*(0,-5,-5).

First component: -2 + 0 = -2.

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

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

Yes, that's correct.

Therefore, the final answer is \boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}.

**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 two given vectors as:

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

We want to find scalars $ x $ and $ y $ such that:

$$
x \mathbf{v}_1 + y \mathbf{v}_2 = \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix}
$$

This gives the system of equations:

$$
\begin{cases}
3x - 5y = -13 \\
x + 2y = 3 \\
2y = 4
\end{cases}
$$

From the third equation: $ y = 2 $

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

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

So, we have:

$$
\begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} = -1 \cdot \mathbf{v}_1 + 2 \cdot \mathbf{v}_2
$$

---

### Step 2: Use linearity of matrix multiplication

Since matrix multiplication is linear, we can write:

$$
\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} = -1 \cdot \mathbf{A} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + 2 \cdot \mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix}
$$

Substitute the known values:

$$
= -1 \cdot \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix} + 2 \cdot \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}
$$

Compute each component:

- First component: $ -1 \cdot 2 + 2 \cdot 0 = -2 $
- Second component: $ -1 \cdot 4 + 2 \cdot (-5) = -4 - 10 = -14 $
- Third component: $ -1 \cdot (-3) + 2 \cdot (-5) = 3 - 10 = -7 $

---

### 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 matrix A is. 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 figure out the matrix A by using these two equations? But wait, matrix A is a 3x3 matrix, right? Because the vectors are in 3-dimensional space. So if I can express the vector (-13, 3, 4) as a linear combination of the vectors (3, 1, 0) and (-5, 2, 2), then I can use the linearity of matrix multiplication to find A times that vector.

But wait, hold on. The problem is that we have two vectors, but we need to express a third vector as a combination of them. However, since we have two vectors in 3D space, they might not span the entire space. So maybe there's another vector we need to consider? Or maybe there's a way to set up equations?

Alternatively, maybe I can set up a system of equations. Let me think. Let me denote the matrix A as:

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

But that's a lot of variables. However, if I can use the given information, maybe I can find some relations between the entries of A.

First, let's consider the first equation: A * [3, 1, 0]^T = [2, 4, -3]^T.

So, multiplying A by the vector (3, 1, 0) gives:

First component: 3a + 1b + 0c = 2
Second component: 3d + 1e + 0f = 4
Third component: 3g + 1h + 0i = -3

So that gives us three equations:

1) 3a + b = 2
2) 3d + e = 4
3) 3g + h = -3

Similarly, the second equation: A * [-5, 2, 2]^T = [0, -5, -5]^T.

So multiplying A by (-5, 2, 2):

First component: -5a + 2b + 2c = 0
Second component: -5d + 2e + 2f = -5
Third component: -5g + 2h + 2i = -5

So that gives us three more equations:

4) -5a + 2b + 2c = 0
5) -5d + 2e + 2f = -5
6) -5g + 2h + 2i = -5

So now, we have six equations, but we have nine variables (a, b, c, d, e, f, g, h, i). So we need more information. But the problem is asking for A times another vector, which is (-13, 3, 4). Maybe we can express (-13, 3, 4) as a linear combination of the two given vectors, and then use linearity?

Let me try that approach. Let me denote the first vector as v1 = [3, 1, 0]^T and the second vector as v2 = [-5, 2, 2]^T. Let me see if I can find scalars x and y such that:

x*v1 + y*v2 = [-13, 3, 4]^T

So, writing that out:

For the first component: 3x -5y = -13
Second component: 1x + 2y = 3
Third component: 0x + 2y = 4

So, let's solve this system. Let me start with the third component equation: 2y = 4 => y = 2.

Then, substitute y = 2 into the second component equation: x + 2*2 = 3 => x + 4 = 3 => x = 3 - 4 = -1.

Now check if these values satisfy the first component equation: 3*(-1) -5*(2) = -3 -10 = -13. Yes! Perfect. So x = -1 and y = 2.

Therefore, the vector [-13, 3, 4]^T is equal to -1*v1 + 2*v2.

Therefore, by linearity of matrix multiplication, A*[-13, 3, 4]^T = -1*A*v1 + 2*A*v2.

We know A*v1 and A*v2 from the problem statement.

So, substituting:

-1*[2, 4, -3]^T + 2*[0, -5, -5]^T

Let me compute each component:

First component: -1*2 + 2*0 = -2 + 0 = -2

Second component: -1*4 + 2*(-5) = -4 -10 = -14

Third component: -1*(-3) + 2*(-5) = 3 -10 = -7

Therefore, the result is [-2, -14, -7]^T.

Wait, but let me check again. Let me recompute:

First component: -1*(2) + 2*(0) = -2 + 0 = -2

Second component: -1*(4) + 2*(-5) = -4 -10 = -14

Third component: -1*(-3) + 2*(-5) = 3 -10 = -7

Yes, that seems correct. So the answer should be the vector (-2, -14, -7).

But let me make sure that I didn't make any mistakes in solving for x and y. Let me check again:

Original equations:

3x -5y = -13

x + 2y = 3

2y = 4 => y = 2. Then x = 3 - 2y = 3 - 4 = -1. Then 3x -5y = 3*(-1) -5*(2) = -3 -10 = -13. Correct.

So that's correct. Therefore, the linear combination is correct.

Therefore, the answer is [-2, -14, -7]^T.

But let me think again. Is there another way to check this? Maybe by trying to find the matrix A? Although that would require more work. Let me try to see if I can find some entries of A.

From the first set of equations:

From equation 1: 3a + b = 2

From equation 4: -5a + 2b + 2c = 0

Similarly, for the second vector, equations 2 and 5:

Equation 2: 3d + e = 4

Equation 5: -5d + 2e + 2f = -5

And for the third component:

Equation 3: 3g + h = -3

Equation 6: -5g + 2h + 2i = -5

So, for the first column, we have equations 1 and 4. Let me try to solve for a, b, c.

From equation 1: 3a + b = 2 => b = 2 - 3a

Substitute into equation 4: -5a + 2*(2 - 3a) + 2c = 0 => -5a + 4 -6a + 2c = 0 => -11a + 4 + 2c = 0 => 2c = 11a -4 => c = (11a -4)/2

But we have three variables here (a, b, c) but only two equations. So we can't determine them uniquely. Similarly for the other columns.

Therefore, the matrix A is not uniquely determined by the given information. However, since we are only asked to compute A times a specific vector, which can be expressed as a linear combination of the given vectors, we don't need to find the entire matrix.

Therefore, our previous approach is valid.

But just to be thorough, let me check if there's any inconsistency. Suppose I pick a value for a, then compute b and c. Let me pick a value for a. Let me choose a = 0. Then b = 2 - 0 = 2. Then c = (0 -4)/2 = -2.

So then, if a = 0, b = 2, c = -2.

Then, let me check if this works with equation 4: -5*0 + 2*2 + 2*(-2) = 0 + 4 -4 = 0. Correct.

So that works.

Similarly, if I pick another value for a, say a = 1. Then b = 2 - 3*1 = -1. Then c = (11*1 -4)/2 = (7)/2 = 3.5.

So then, equation 4: -5*1 + 2*(-1) + 2*(3.5) = -5 -2 +7 = 0. Correct.

So, there are infinitely many solutions for the first column. However, when we apply A to the vector [-13, 3, 4], which is a combination of v1 and v2, the result only depends on the action of A on v1 and v2, which are given. Therefore, regardless of the other entries of A, the result is determined.

Therefore, our answer is correct.

Alternatively, if I wanted to find the entire matrix A, I would need more information. But since we don't have it, and the problem only requires a specific application of A, our method is valid.

Therefore, the answer is the vector (-2, -14, -7).

But let me check once more by another approach. Suppose I try to find A. Let me think.

Since we have two vectors, maybe we can find A by setting up equations. However, since we have two vectors, we can't determine A uniquely. However, if we assume that the third column is arbitrary? Wait, but the problem is that the vector [-13, 3, 4] is in the span of v1 and v2, so even if we don't know the third column, since we are not using it in the calculation, it's okay.

Alternatively, maybe I can think of A as a linear transformation, and since we know its action on two vectors, we can find its action on any vector in their span. Since [-13, 3, 4] is in their span, we can compute it as we did.

Therefore, I think the answer is correct.

Wait, but let me check once again the calculation of -1*(2,4,-3) + 2*(0,-5,-5).

First component: -2 + 0 = -2.

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

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

Yes, that's correct.

Therefore, the final answer is \boxed{\begin{pmatrix} -2 \\ -14 \\ -7 \end{pmatrix}}.

**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 two given vectors as:

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

We want to find scalars $ x $ and $ y $ such that:

$$
x \mathbf{v}_1 + y \mathbf{v}_2 = \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix}
$$

This gives the system of equations:

$$
\begin{cases}
3x - 5y = -13 \\
x + 2y = 3 \\
2y = 4
\end{cases}
$$

From the third equation: $ y = 2 $

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

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

So, we have:

$$
\begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} = -1 \cdot \mathbf{v}_1 + 2 \cdot \mathbf{v}_2
$$

---

### Step 2: Use linearity of matrix multiplication

Since matrix multiplication is linear, we can write:

$$
\mathbf{A} \begin{pmatrix} -13 \\ 3 \\ 4 \end{pmatrix} = -1 \cdot \mathbf{A} \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + 2 \cdot \mathbf{A} \begin{pmatrix} -5 \\ 2 \\ 2 \end{pmatrix}
$$

Substitute the known values:

$$
= -1 \cdot \begin{pmatrix} 2 \\ 4 \\ -3 \end{pmatrix} + 2 \cdot \begin{pmatrix} 0 \\ -5 \\ -5 \end{pmatrix}
$$

Compute each component:

- First component: $ -1 \cdot 2 + 2 \cdot 0 = -2 $
- Second component: $ -1 \cdot 4 + 2 \cdot (-5) = -4 - 10 = -14 $
- Third component: $ -1 \cdot (-3) + 2 \cdot (-5) = 3 - 10 = -7 $

---

### 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 down:

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

To find the LCM, I take the highest power of each prime number that appears in the factorizations. So for 2, the highest power is 2² from 4. For 3, it's 3¹, and for 5, it's 5¹. Multiplying those together gives me the LCM.

So LCM = 2² × 3 × 5. Let me calculate that. 2² is 4, times 3 is 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.

Therefore, the problem reduces to finding how many multiples of 60 are there in the first 100 positive integers. Let me think. The multiples of 60 are 60, 120, 180, etc. But since we're only going up to 100, the first multiple is 60, and the next one would be 120, which is already over 100. So there's only one multiple of 60 within the first 100 numbers.

Wait, but let me double-check. Let me list them out. Starting from 1:

1×60 = 60

2×60 = 120, which is 120, which is more than 100. So yes, only 60 is the multiple of 60 within 1-100. Therefore, the answer should be 1?

But wait, hold on. Let me make sure I didn't make a mistake in calculating the LCM. Maybe I should check if 60 is indeed divisible by 3, 4, and 5.

Dividing 60 by 3: 60 ÷ 3 = 20, which is an integer. Dividing by 4: 60 ÷ 4 = 15, which is also an integer. Dividing by 5: 60 ÷ 5 = 12, which is an integer. So yes, 60 is divisible by all three. So that's correct.

But let me think again. Is there another number between 1 and 100 that is divisible by 3, 4, and 5? For example, 60 is the first one. The next one would be 120, which is outside the range. So there's only one such number. Therefore, the answer is 1?

But wait, maybe I need to check if there are numbers that are divisible by 3, 4, and 5 without necessarily being multiples of 60? But no, if a number is divisible by 3, 4, and 5, then it must be divisible by their LCM. Because LCM is the least common multiple, so any common multiple must be a multiple of the LCM. Therefore, there can't be any other numbers besides multiples of 60 that are divisible by all three. Therefore, only 60 is in the range 1-100. Therefore, the answer is 1.

But let me check with another approach. Let me think about divisibility rules. For a number to be divisible by 3, the sum of its digits must be divisible by 3. For 4, the last two digits must form a number that's divisible by 4. For 5, the last digit must be 0 or 5. So if a number is divisible by 5, it must end with 0 or 5. But if it's also divisible by 4, then the last two digits must form a number divisible by 4. Let me think. If the number ends with 5, then the last two digits would be something like x5. But numbers ending with 5 are odd, and 4 divides only even numbers. Therefore, if a number ends with 5, it can't be divisible by 4. Therefore, the number must end with 0. So the number must end with 0. Therefore, it's divisible by 10. So the number must be divisible by 10 and also by 3 and 4. Wait, but if it's divisible by 10, then it's divisible by 2 and 5. But since we already have divisibility by 4, which is a higher power of 2. So let me think.

So if a number is divisible by 4 and 5, then it must be divisible by LCM(4,5) which is 20. Then, since it's also divisible by 3, it must be divisible by LCM(20,3) which is 60. Therefore, same conclusion as before. Therefore, numbers divisible by 3,4,5 are exactly the numbers divisible by 60. Therefore, only 60 in 1-100. Therefore, answer is 1.

But let me check if there are numbers between 1-100 divisible by 60. 60 is 60, 120 is 120. So only 60. Therefore, answer is 1. But wait, maybe I made a mistake here? Let me check numbers divisible by 3, 4, and 5. Let me check 60. 60 divided by 3 is 20, 60 divided by 4 is 15, 60 divided by 5 is 12. All integers. What about 120? 120 divided by 3 is 40, 4 is 30, 5 is 24. But 120 is beyond 100. So yes, only 60. Therefore, answer is 1. So maybe the answer is 1?

But wait, let me check another way. Let me list numbers divisible by 3, 4, and 5. Let me think of numbers divisible by 3 and 4 first. The LCM of 3 and 4 is 12. So numbers divisible by 12. Then, numbers divisible by 12 and 5. So LCM of 12 and 5 is 60. Therefore, numbers divisible by 60. So same conclusion. Therefore, numbers divisible by 3,4,5 are numbers divisible by 60. Therefore, only 60 in 1-100. Therefore, answer is 1.

But maybe I need to check if there are numbers that are divisible by 3, 4, and 5 but not by 60? But that's impossible because LCM(3,4,5)=60. So any number divisible by all three must be divisible by 60. Therefore, there are no such numbers outside multiples of 60. Therefore, only 60. Therefore, answer is 1.

But wait, let me check with actual numbers. Let me check numbers from 1 to 100. Let me check if there are numbers divisible by 3, 4, and 5. Let me check 60. Yes. What about 30? 30 is divisible by 3 and 5, but 30 divided by 4 is 7.5, which is not integer. So no. How about 40? Divided by 3 is 13.333... Not integer. 20? Divided by 3 is 6.666... No. 15? Divided by 4 is 3.75. No. 12? Divided by 5 is 2.4. No. 24? Divided by 5 is 4.8. No. 36? Divided by 5 is 7.2. No. 48? Divided by 5 is 9.6. No. 60? Divided by 5 is 12. Yes. Then 72? Divided by 5 is 14.4. No. 84? Divided by 5 is 16.8. No. 96? Divided by 5 is 19.2. No. So only 60. Therefore, answer is 1. Therefore, the answer is 1. So I think that's correct.

But wait, maybe I need to check if there's another number. Let me think. Let me check 0? But 0 is not in the first 100 positive integers. The first one is 1. So 0 is excluded. Therefore, only 60. Therefore, answer is 1. Therefore, I think the answer is 1. But let me check once again.

Alternatively, maybe I can use the formula for the number of multiples. The number of multiples of a number n in the range 1 to N is floor(N/n). So here, N is 100, and n is 60. Therefore, floor(100/60) = floor(1.666...) = 1. Therefore, 1. Therefore, answer is 1. Therefore, that's another way to get the same answer. Therefore, I think that's correct.

But maybe I made a mistake in calculating LCM? Let me check again. LCM of 3,4,5. Let me think. Prime factors:

- 3: 3
- 4: 2²
- 5: 5

So LCM is 2² * 3 * 5 = 4 * 3 * 5 = 60. Correct. So that's right. Therefore, answer is 1. Therefore, I think the answer is 1. Therefore, the number of integers between 1 and 100 divisible by 3,4,5 is 1. Therefore, the answer is \boxed{1}.

Wait, but hold on. Let me check once again. Maybe I need to check if there are numbers that are divisible by 3, 4, and 5 but not by 60? For example, suppose there is a number that is divisible by 3, 4, and 5 but not by 60. But since LCM(3,4,5) is 60, that would mean that 60 is the least common multiple, so any common multiple must be a multiple of 60. Therefore, there can't be such a number. Therefore, only multiples of 60. Therefore, only 60 in 1-100. Therefore, answer is 1. Therefore, I think that's correct.

Alternatively, maybe I can think of it as follows: For a number to be divisible by 3, 4, and 5, it must be divisible by each of them. So, let me check numbers divisible by 3 and 4 first. As before, LCM(3,4)=12. So numbers divisible by 12. Then, among those numbers, which are also divisible by 5. So numbers divisible by 12 and 5. Which is LCM(12,5)=60. Therefore, numbers divisible by 60. Therefore, same conclusion. Therefore, answer is 1. Therefore, I think that's correct.

Alternatively, maybe I can think of it as follows: Let me list numbers divisible by 5 between 1 and 100. Those are 5,10,15,...,100. Then, among those numbers, which are also divisible by 4 and 3. So first, numbers divisible by 5 and 4. Which is LCM(5,4)=20. So numbers divisible by 20. Then, among those numbers, which are also divisible by 3. So LCM(20,3)=60. Therefore, numbers divisible by 60. Therefore, same conclusion. Therefore, answer is 1. Therefore, I think that's correct.

Therefore, after multiple approaches and checking, I think the answer is 1. Therefore, the number of integers between 1 and 100 divisible by 3,4,5 is 1. Therefore, the answer is \boxed{1}.

**Final Answer**
\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

We are looking for numbers that are divisible by **3**, **4**, and **5**. That means the number must be a **common multiple** of all three. The **least common multiple (LCM)** of these numbers will give us 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 in the range from **1 to 100**.

- The first multiple of 60 is $ 60 \times 1 = 60 $
- The next multiple is $ 60 \times 2 = 120 $, which is **greater than 100**

So, the only multiple of 60 in the range is **60**.

---

### Step 4: Final Answer

Therefore, there is **only one** number between 1 and 100 that is divisible by **3, 4, and 5**.

$$
\boxed{1}
$$

[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 down:

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

To find the LCM, I take the highest power of each prime number that appears in the factorizations. So for 2, the highest power is 2² from 4. For 3, it's 3¹, and for 5, it's 5¹. Multiplying those together gives me the LCM.

So LCM = 2² × 3 × 5. Let me calculate that. 2² is 4, times 3 is 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.

Therefore, the problem reduces to finding how many multiples of 60 are there in the first 100 positive integers. Let me think. The multiples of 60 are 60, 120, 180, etc. But since we're only going up to 100, the first multiple is 60, and the next one would be 120, which is already over 100. So there's only one multiple of 60 within the first 100 numbers.

Wait, but let me double-check. Let me list them out. Starting from 1:

1×60 = 60

2×60 = 120, which is 120, which is more than 100. So yes, only 60 is the multiple of 60 within 1-100. Therefore, the answer should be 1?

But wait, hold on. Let me make sure I didn't make a mistake in calculating the LCM. Maybe I should check if 60 is indeed divisible by 3, 4, and 5.

Dividing 60 by 3: 60 ÷ 3 = 20, which is an integer. Dividing by 4: 60 ÷ 4 = 15, which is also an integer. Dividing by 5: 60 ÷ 5 = 12, which is an integer. So yes, 60 is divisible by all three. So that's correct.

But let me think again. Is there another number between 1 and 100 that is divisible by 3, 4, and 5? For example, 60 is the first one. The next one would be 120, which is outside the range. So there's only one such number. Therefore, the answer is 1?

But wait, maybe I need to check if there are numbers that are divisible by 3, 4, and 5 without necessarily being multiples of 60? But no, if a number is divisible by 3, 4, and 5, then it must be divisible by their LCM. Because LCM is the least common multiple, so any common multiple must be a multiple of the LCM. Therefore, there can't be any other numbers besides multiples of 60 that are divisible by all three. Therefore, only 60 is in the range 1-100. Therefore, the answer is 1.

But let me check with another approach. Let me think about divisibility rules. For a number to be divisible by 3, the sum of its digits must be divisible by 3. For 4, the last two digits must form a number that's divisible by 4. For 5, the last digit must be 0 or 5. So if a number is divisible by 5, it must end with 0 or 5. But if it's also divisible by 4, then the last two digits must form a number divisible by 4. Let me think. If the number ends with 5, then the last two digits would be something like x5. But numbers ending with 5 are odd, and 4 divides only even numbers. Therefore, if a number ends with 5, it can't be divisible by 4. Therefore, the number must end with 0. So the number must end with 0. Therefore, it's divisible by 10. So the number must be divisible by 10 and also by 3 and 4. Wait, but if it's divisible by 10, then it's divisible by 2 and 5. But since we already have divisibility by 4, which is a higher power of 2. So let me think.

So if a number is divisible by 4 and 5, then it must be divisible by LCM(4,5) which is 20. Then, since it's also divisible by 3, it must be divisible by LCM(20,3) which is 60. Therefore, same conclusion as before. Therefore, numbers divisible by 3,4,5 are exactly the numbers divisible by 60. Therefore, only 60 in 1-100. Therefore, answer is 1.

But let me check if there are numbers between 1-100 divisible by 60. 60 is 60, 120 is 120. So only 60. Therefore, answer is 1. But wait, maybe I made a mistake here? Let me check numbers divisible by 3, 4, and 5. Let me check 60. 60 divided by 3 is 20, 60 divided by 4 is 15, 60 divided by 5 is 12. All integers. What about 120? 120 divided by 3 is 40, 4 is 30, 5 is 24. But 120 is beyond 100. So yes, only 60. Therefore, answer is 1. So maybe the answer is 1?

But wait, let me check another way. Let me list numbers divisible by 3, 4, and 5. Let me think of numbers divisible by 3 and 4 first. The LCM of 3 and 4 is 12. So numbers divisible by 12. Then, numbers divisible by 12 and 5. So LCM of 12 and 5 is 60. Therefore, numbers divisible by 60. So same conclusion. Therefore, numbers divisible by 3,4,5 are numbers divisible by 60. Therefore, only 60 in 1-100. Therefore, answer is 1.

But maybe I need to check if there are numbers that are divisible by 3, 4, and 5 but not by 60? But that's impossible because LCM(3,4,5)=60. So any number divisible by all three must be divisible by 60. Therefore, there are no such numbers outside multiples of 60. Therefore, only 60. Therefore, answer is 1.

But wait, let me check with actual numbers. Let me check numbers from 1 to 100. Let me check if there are numbers divisible by 3, 4, and 5. Let me check 60. Yes. What about 30? 30 is divisible by 3 and 5, but 30 divided by 4 is 7.5, which is not integer. So no. How about 40? Divided by 3 is 13.333... Not integer. 20? Divided by 3 is 6.666... No. 15? Divided by 4 is 3.75. No. 12? Divided by 5 is 2.4. No. 24? Divided by 5 is 4.8. No. 36? Divided by 5 is 7.2. No. 48? Divided by 5 is 9.6. No. 60? Divided by 5 is 12. Yes. Then 72? Divided by 5 is 14.4. No. 84? Divided by 5 is 16.8. No. 96? Divided by 5 is 19.2. No. So only 60. Therefore, answer is 1. Therefore, the answer is 1. So I think that's correct.

But wait, maybe I need to check if there's another number. Let me think. Let me check 0? But 0 is not in the first 100 positive integers. The first one is 1. So 0 is excluded. Therefore, only 60. Therefore, answer is 1. Therefore, I think the answer is 1. But let me check once again.

Alternatively, maybe I can use the formula for the number of multiples. The number of multiples of a number n in the range 1 to N is floor(N/n). So here, N is 100, and n is 60. Therefore, floor(100/60) = floor(1.666...) = 1. Therefore, 1. Therefore, answer is 1. Therefore, that's another way to get the same answer. Therefore, I think that's correct.

But maybe I made a mistake in calculating LCM? Let me check again. LCM of 3,4,5. Let me think. Prime factors:

- 3: 3
- 4: 2²
- 5: 5

So LCM is 2² * 3 * 5 = 4 * 3 * 5 = 60. Correct. So that's right. Therefore, answer is 1. Therefore, I think the answer is 1. Therefore, the number of integers between 1 and 100 divisible by 3,4,5 is 1. Therefore, the answer is \boxed{1}.

Wait, but hold on. Let me check once again. Maybe I need to check if there are numbers that are divisible by 3, 4, and 5 but not by 60? For example, suppose there is a number that is divisible by 3, 4, and 5 but not by 60. But since LCM(3,4,5) is 60, that would mean that 60 is the least common multiple, so any common multiple must be a multiple of 60. Therefore, there can't be such a number. Therefore, only multiples of 60. Therefore, only 60 in 1-100. Therefore, answer is 1. Therefore, I think that's correct.

Alternatively, maybe I can think of it as follows: For a number to be divisible by 3, 4, and 5, it must be divisible by each of them. So, let me check numbers divisible by 3 and 4 first. As before, LCM(3,4)=12. So numbers divisible by 12. Then, among those numbers, which are also divisible by 5. So numbers divisible by 12 and 5. Which is LCM(12,5)=60. Therefore, numbers divisible by 60. Therefore, same conclusion. Therefore, answer is 1. Therefore, I think that's correct.

Alternatively, maybe I can think of it as follows: Let me list numbers divisible by 5 between 1 and 100. Those are 5,10,15,...,100. Then, among those numbers, which are also divisible by 4 and 3. So first, numbers divisible by 5 and 4. Which is LCM(5,4)=20. So numbers divisible by 20. Then, among those numbers, which are also divisible by 3. So LCM(20,3)=60. Therefore, numbers divisible by 60. Therefore, same conclusion. Therefore, answer is 1. Therefore, I think that's correct.

Therefore, after multiple approaches and checking, I think the answer is 1. Therefore, the number of integers between 1 and 100 divisible by 3,4,5 is 1. Therefore, the answer is \boxed{1}.

**Final Answer**
\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

We are looking for numbers that are divisible by **3**, **4**, and **5**. That means the number must be a **common multiple** of all three. The **least common multiple (LCM)** of these numbers will give us 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 in the range from **1 to 100**.

- The first multiple of 60 is $ 60 \times 1 = 60 $
- The next multiple is $ 60 \times 2 = 120 $, which is **greater than 100**

So, the only multiple of 60 in the range is **60**.

---

### Step 4: Final Answer

Therefore, there is **only one** number between 1 and 100 that is divisible by **3, 4, and 5**.

$$
\boxed{1}
$$

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. Hmm.

First, let me break down the information. Total questions: 100. He answered 80, so that means he left 20 questions unanswered. But the score comes from the answered ones: correct ones give +5, incorrect ones give -2. The unanswered ones don't affect the score. So, the total score is based on the 80 questions he answered.

Let me define variables. Let me let x be the number of correct answers. Then, since he answered 80 questions, the number of incorrect answers would be 80 - x. Because if he answered 80 questions, and x are correct, the rest must be incorrect.

So, his total score would be: (number of correct answers * 5) + (number of incorrect answers * -2). Which is 5x - 2*(80 - x). And this equals 232 points. So I can set up the equation:

5x - 2(80 - x) = 232

Let me write that out:

5x - 2*(80 - x) = 232

Okay, let me solve this equation step by step. First, distribute the -2 into the parentheses:

5x - 160 + 2x = 232

Wait, because -2*(80 - x) is -160 + 2x. So combining like terms:

5x + 2x - 160 = 232

Which is 7x - 160 = 232

Then, add 160 to both sides:

7x = 232 + 160

Let me calculate 232 + 160. 200 + 100 is 300, 32 + 60 is 92, so total is 392? Wait, 232 + 160. Let me check again:

232 + 160. 200 + 100 is 300, 32 + 60 is 92. So 300 + 92 is 392. Yes. So 7x = 392.

Then, divide both sides by 7:

x = 392 / 7

Let me compute that. 7*50 is 350. 392 - 350 is 42. 42 / 7 is 6. So total is 50 + 6 = 56. So x = 56.

So Frank answered 56 questions correctly. Let me check if this makes sense.

If he got 56 correct, then incorrect answers are 80 - 56 = 24. So his score would be 56*5 - 24*2. Let me calculate that.

56*5: 50*5=250, 6*5=30, so 250+30=280. Then 24*2=48. So 280 - 48 = 232. Which matches the given score. So that seems correct.

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

Alternatively, maybe using the total number of questions. But since he answered 80, maybe not necessary. But let me think.

Alternatively, suppose we let c be correct, i be incorrect, and u be unanswered. Then:

c + i + u = 100

But he answered 80 questions, so c + i = 80, and u = 20.

His score is 5c - 2i = 232.

So we have two equations:

1) c + i = 80

2) 5c - 2i = 232

So we can solve this system. Let me try solving it.

From equation 1, i = 80 - c. Substitute into equation 2:

5c - 2*(80 - c) = 232

Which is exactly the same equation as before. So solving gives c = 56. So same result. So that's consistent.

Alternatively, maybe think in terms of total points. But I think that's the same approach.

Alternatively, think about how many points he would have if all answers were correct, then adjust for incorrect ones. Let me try that.

If all 80 answered questions were correct, he would have 80*5 = 400 points. But he has 232 points. The difference is 400 - 232 = 168 points. Each incorrect answer is not only not giving 5 points but also subtracting 2 points. So compared to a correct answer, an incorrect answer is worth 5 - (-2) = 7 points less? Wait, maybe not. Let me think.

Wait, if you have a correct answer, you get +5. If you have an incorrect answer, you get -2. So the difference between a correct and incorrect answer is 5 - (-2) = 7 points. So if you have x correct answers and (80 - x) incorrect answers, then compared to all correct, you have (80 - x) incorrect answers which each are 7 points less. Wait, but how does that translate?

Wait, total score is 5x - 2(80 - x). If all were correct, it would be 5*80 = 400. So the difference between that and his actual score is 400 - 232 = 168. Which is equal to the total loss due to incorrect answers. But each incorrect answer is not just losing 5 points (since it's not correct), but also subtracting 2. So maybe the total loss is 5*(number of incorrect answers) + 2*(number of incorrect answers)? Wait, no.

Wait, if you have an incorrect answer, instead of gaining 5, you lose 2. So the difference is 5 - (-2) = 7 points less. So for each incorrect answer, compared to a correct one, you have 7 points less. Therefore, if there are (80 - x) incorrect answers, the total loss is 7*(80 - x). Therefore, total score would be 400 - 7*(80 - x) = 232. Let me check this.

So:

400 - 7*(80 - x) = 232

Let me solve this:

-7*(80 - x) = 232 - 400 = -168

Divide both sides by -7:

80 - x = (-168)/(-7) = 24

Therefore, 80 - x = 24 => x = 80 - 24 = 56. Same answer. So that works too. So that's another way to think about it.

So regardless of the method, the answer is 56. So I think that's correct.

But let me check once more with actual numbers. If he got 56 correct, 24 incorrect, 20 unanswered.

Total score: 56*5 = 280, 24*(-2) = -48. 280 - 48 = 232. Correct. So that's exactly the score given. So that's correct.

Therefore, the answer is 56. So I think that's solid.

**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:
- $ x $ = number of correct answers
- $ 80 - x $ = number of incorrect answers (since he answered 80 questions in total)
- Unanswered questions = 20 (since 100 - 80 = 20)

---

### Step 2: Set Up the Equation

Frank's total score is based only on the 80 answered questions. The score is calculated as:

$$
\text{Score} = 5x - 2(80 - x)
$$

We are told his score is 232, so:

$$
5x - 2(80 - x) = 232
$$

---

### Step 3: Solve the Equation

Distribute the -2:

$$
5x - 160 + 2x = 232
$$

Combine like terms:

$$
7x - 160 = 232
$$

Add 160 to both sides:

$$
7x = 392
$$

Divide both sides by 7:

$$
x = \frac{392}{7} = 56
$$

---

### Step 4: Verify the Solution

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

This matches the given 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. Hmm.

First, let me break down the information. Total questions: 100. He answered 80, so that means he left 20 questions unanswered. But the score comes from the answered ones: correct ones give +5, incorrect ones give -2. The unanswered ones don't affect the score. So, the total score is based on the 80 questions he answered.

Let me define variables. Let me let x be the number of correct answers. Then, since he answered 80 questions, the number of incorrect answers would be 80 - x. Because if he answered 80 questions, and x are correct, the rest must be incorrect.

So, his total score would be: (number of correct answers * 5) + (number of incorrect answers * -2). Which is 5x - 2*(80 - x). And this equals 232 points. So I can set up the equation:

5x - 2(80 - x) = 232

Let me write that out:

5x - 2*(80 - x) = 232

Okay, let me solve this equation step by step. First, distribute the -2 into the parentheses:

5x - 160 + 2x = 232

Wait, because -2*(80 - x) is -160 + 2x. So combining like terms:

5x + 2x - 160 = 232

Which is 7x - 160 = 232

Then, add 160 to both sides:

7x = 232 + 160

Let me calculate 232 + 160. 200 + 100 is 300, 32 + 60 is 92, so total is 392? Wait, 232 + 160. Let me check again:

232 + 160. 200 + 100 is 300, 32 + 60 is 92. So 300 + 92 is 392. Yes. So 7x = 392.

Then, divide both sides by 7:

x = 392 / 7

Let me compute that. 7*50 is 350. 392 - 350 is 42. 42 / 7 is 6. So total is 50 + 6 = 56. So x = 56.

So Frank answered 56 questions correctly. Let me check if this makes sense.

If he got 56 correct, then incorrect answers are 80 - 56 = 24. So his score would be 56*5 - 24*2. Let me calculate that.

56*5: 50*5=250, 6*5=30, so 250+30=280. Then 24*2=48. So 280 - 48 = 232. Which matches the given score. So that seems correct.

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

Alternatively, maybe using the total number of questions. But since he answered 80, maybe not necessary. But let me think.

Alternatively, suppose we let c be correct, i be incorrect, and u be unanswered. Then:

c + i + u = 100

But he answered 80 questions, so c + i = 80, and u = 20.

His score is 5c - 2i = 232.

So we have two equations:

1) c + i = 80

2) 5c - 2i = 232

So we can solve this system. Let me try solving it.

From equation 1, i = 80 - c. Substitute into equation 2:

5c - 2*(80 - c) = 232

Which is exactly the same equation as before. So solving gives c = 56. So same result. So that's consistent.

Alternatively, maybe think in terms of total points. But I think that's the same approach.

Alternatively, think about how many points he would have if all answers were correct, then adjust for incorrect ones. Let me try that.

If all 80 answered questions were correct, he would have 80*5 = 400 points. But he has 232 points. The difference is 400 - 232 = 168 points. Each incorrect answer is not only not giving 5 points but also subtracting 2 points. So compared to a correct answer, an incorrect answer is worth 5 - (-2) = 7 points less? Wait, maybe not. Let me think.

Wait, if you have a correct answer, you get +5. If you have an incorrect answer, you get -2. So the difference between a correct and incorrect answer is 5 - (-2) = 7 points. So if you have x correct answers and (80 - x) incorrect answers, then compared to all correct, you have (80 - x) incorrect answers which each are 7 points less. Wait, but how does that translate?

Wait, total score is 5x - 2(80 - x). If all were correct, it would be 5*80 = 400. So the difference between that and his actual score is 400 - 232 = 168. Which is equal to the total loss due to incorrect answers. But each incorrect answer is not just losing 5 points (since it's not correct), but also subtracting 2. So maybe the total loss is 5*(number of incorrect answers) + 2*(number of incorrect answers)? Wait, no.

Wait, if you have an incorrect answer, instead of gaining 5, you lose 2. So the difference is 5 - (-2) = 7 points less. So for each incorrect answer, compared to a correct one, you have 7 points less. Therefore, if there are (80 - x) incorrect answers, the total loss is 7*(80 - x). Therefore, total score would be 400 - 7*(80 - x) = 232. Let me check this.

So:

400 - 7*(80 - x) = 232

Let me solve this:

-7*(80 - x) = 232 - 400 = -168

Divide both sides by -7:

80 - x = (-168)/(-7) = 24

Therefore, 80 - x = 24 => x = 80 - 24 = 56. Same answer. So that works too. So that's another way to think about it.

So regardless of the method, the answer is 56. So I think that's correct.

But let me check once more with actual numbers. If he got 56 correct, 24 incorrect, 20 unanswered.

Total score: 56*5 = 280, 24*(-2) = -48. 280 - 48 = 232. Correct. So that's exactly the score given. So that's correct.

Therefore, the answer is 56. So I think that's solid.

**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:
- $ x $ = number of correct answers
- $ 80 - x $ = number of incorrect answers (since he answered 80 questions in total)
- Unanswered questions = 20 (since 100 - 80 = 20)

---

### Step 2: Set Up the Equation

Frank's total score is based only on the 80 answered questions. The score is calculated as:

$$
\text{Score} = 5x - 2(80 - x)
$$

We are told his score is 232, so:

$$
5x - 2(80 - x) = 232
$$

---

### Step 3: Solve the Equation

Distribute the -2:

$$
5x - 160 + 2x = 232
$$

Combine like terms:

$$
7x - 160 = 232
$$

Add 160 to both sides:

$$
7x = 392
$$

Divide both sides by 7:

$$
x = \frac{392}{7} = 56
$$

---

### Step 4: Verify the Solution

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

This matches the given 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.