Explanation:
Okay, so I need to find p(0) + p(4) where p(x) is a monic polynomial of degree 4, and three of its roots are 1, 2, and 3. Let me start by recalling what a monic polynomial is. A monic polynomial is one where the leading coefficient (the coefficient of the highest power of x) is 1. Since it's a degree 4 polynomial, it has four roots (counting multiplicities). They told me three of the roots are 1, 2, and 3, so there must be a fourth root. Let me call that fourth root 'a'.

So, the polynomial can be written in factored form as p(x) = (x - 1)(x - 2)(x - 3)(x - a). Since it's monic, the leading coefficient is 1, which matches this form. Now, I need to figure out what 'a' is. Wait, but the problem doesn't give me any other information about the polynomial. Hmm. Maybe I don't need to find 'a' explicitly? Let me think.

The question asks for p(0) + p(4). Let me compute each term separately. Let's start with p(0). If I plug in x = 0 into the polynomial, I get p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a). Let me compute that. Multiplying the constants first: (-1)(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. So p(0) = 6a.

Now, let's compute p(4). Plugging in x = 4, we get p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a). Multiplying those constants: 3*2 is 6, 6*1 is 6, so p(4) = 6*(4 - a) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Wait, that's interesting. The 6a and -6a cancel each other out, leaving just 24. So regardless of the value of 'a', p(0) + p(4) is always 24? That seems surprising, but maybe that's correct. Let me check my steps again to make sure I didn't make a mistake.

First, p(x) is monic quartic with roots 1, 2, 3, and a. So factored form is correct. Then p(0) is product of (-1)(-2)(-3)(-a). Let me compute that again. (-1)*(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. Yes, that's correct. Then p(4) is (3)(2)(1)(4 - a) which is 6*(4 - a) = 24 - 6a. Then adding them together: 6a + 24 - 6a = 24. So yes, the answer is 24.

But wait, is there a possibility that the polynomial has a repeated root? The problem says "three of the roots are 1, 2, and 3". It doesn't specify if they are distinct or not. But since it's a degree 4 polynomial, and three roots are given, the fourth root could be any number, possibly one of the existing roots. However, even if there is a repeated root, the calculation for p(0) + p(4) would still hold because the fourth root is still 'a', whether it's equal to 1, 2, 3, or something else. For example, if a = 1, then p(x) would be (x - 1)^2(x - 2)(x - 3). But in that case, p(0) would be (-1)^2*(-2)*(-3) = 1*6 = 6, and p(4) would be (3)^2*(2)*(1) = 9*2 = 18. Then p(0) + p(4) = 6 + 18 = 24. Same result. Similarly, if a = 2, p(0) would be (-1)(-2)^2*(-3)(-2) = (-1)*(4)*(-3)*(-2). Wait, let me compute that again. Wait, if a = 2, then p(x) = (x - 1)(x - 2)^2(x - 3). Then p(0) = (-1)(-2)^2(-3) = (-1)*(4)*(-3) = 12. Then p(4) = (3)(2)^2(1) = 3*4*1 = 12. Then p(0) + p(4) = 12 + 12 = 24. Same result. Similarly, if a = 3, p(0) would be (-1)(-2)(-3)^2(-a) = (-1)(-2)(9)(-3) = (-1)(-2) is 2, 2*9 is 18, 18*(-3) is -54? Wait, hold on. Wait, if a = 3, then p(x) = (x - 1)(x - 2)(x - 3)^2. Then p(0) = (-1)(-2)(-3)^2 = (-1)(-2)(9) = 2*9 = 18. Then p(4) = (3)(2)(1)^2 = 3*2*1 = 6. Then p(0) + p(4) = 18 + 6 = 24. Still 24. So even if there are repeated roots, the answer is still 24. Therefore, regardless of the value of 'a', p(0) + p(4) is 24. Therefore, the answer is 24.

But let me think again. Is there a way that the problem could have more constraints? For example, maybe the polynomial is supposed to have distinct roots? But the problem didn't specify that. It just said three of the roots are 1, 2, 3. So the fourth root could be anything. Therefore, regardless of that, the answer is 24.

Alternatively, maybe there's another approach. Let me try expanding the polynomial. Let me write p(x) = (x - 1)(x - 2)(x - 3)(x - a). Let me first compute (x - 1)(x - 2)(x - 3). Let me compute that first. Let me compute (x - 1)(x - 2) first. That is x^2 - 3x + 2. Then multiply by (x - 3): (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Wait, let me do that step by step. Multiply x^2 by (x - 3): x^3 - 3x^2. Then -3x times (x - 3): -3x^2 + 9x. Then 2 times (x - 3): 2x - 6. Now add them all together: x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Combine like terms: x^3 - 6x^2 + 11x - 6. So (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6. Therefore, p(x) = (x^3 - 6x^2 + 11x - 6)(x - a). Let me expand this. Multiply each term by (x - a):

First, x^3*(x - a) = x^4 - a x^3.

Then -6x^2*(x - a) = -6x^3 + 6a x^2.

Then 11x*(x - a) = 11x^2 - 11a x.

Then -6*(x - a) = -6x + 6a.

Now combine all these terms:

x^4 - a x^3 - 6x^3 + 6a x^2 + 11x^2 - 11a x - 6x + 6a.

Combine like terms:

x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

So p(x) = x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

Now, if I need to compute p(0) + p(4), let me compute p(0) first. p(0) is the constant term, which is 6a. Then p(4) is the value of the polynomial at x = 4. Let me compute p(4):

p(4) = (4)^4 + (-a - 6)(4)^3 + (6a + 11)(4)^2 + (-11a - 6)(4) + 6a.

Compute each term:

4^4 = 256.

(-a - 6)(4)^3 = (-a - 6)(64) = -64a - 384.

(6a + 11)(4)^2 = (6a + 11)(16) = 96a + 176.

(-11a - 6)(4) = -44a - 24.

And the last term is 6a.

So adding all these together:

256 + (-64a - 384) + (96a + 176) + (-44a - 24) + 6a.

Let me combine like terms step by step.

First, constants: 256 - 384 + 176 - 24.

Compute 256 - 384: that's -128.

Then -128 + 176: that's 48.

Then 48 - 24: that's 24.

Now, the terms with 'a':

-64a + 96a - 44a + 6a.

Compute -64a + 96a = 32a.

32a - 44a = -12a.

-12a + 6a = -6a.

So overall, p(4) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Same result as before. So regardless of the value of 'a', the answer is 24. Therefore, the answer is 24.

But let me check once more with a specific example. Let me choose a specific value for 'a' and compute p(0) + p(4). Let's say a = 0. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 0) = x(x - 1)(x - 2)(x - 3). Then p(0) = 0, since there's an x term. Then p(4) = 4*(4 - 1)(4 - 2)(4 - 3) = 4*3*2*1 = 24. Therefore, p(0) + p(4) = 0 + 24 = 24. Correct.

Another example: let a = 4. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 4). Then p(0) = (-1)(-2)(-3)(-4) = 24. p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - 4) = 3*2*1*0 = 0. Therefore, p(0) + p(4) = 24 + 0 = 24. Correct again.

Another example: let a = 5. Then p(0) = (-1)(-2)(-3)(-5) = 30. p(4) = (3)(2)(1)(-1) = -6. Then p(0) + p(4) = 30 - 6 = 24. Correct.

So regardless of the value of 'a', the answer is always 24. Therefore, the answer is 24.

But wait, the problem says "three of the roots of p(x) are 1, 2, and 3". So maybe there's a possibility that the fourth root is something else? But even so, as shown, the answer is always 24. Therefore, the answer is 24.

Alternatively, maybe there's a different approach. Let me think. Since p(x) is a monic quartic with roots 1, 2, 3, and a. Then p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) is the product of the roots multiplied by (-1)^4, since for a monic polynomial, the constant term is (-1)^n times the product of the roots, where n is the degree. Wait, for a monic polynomial of degree 4, p(x) = x^4 + ... + (-1)^4 * product of roots. Wait, actually, the constant term is (-1)^4 times the product of the roots? Wait, no. Wait, the constant term is the product of the roots multiplied by (-1)^degree. Wait, let me recall Vieta's formula. For a monic polynomial p(x) = x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0, the product of the roots is c_0 * (-1)^4 = c_0. Wait, Vieta's formula says that the product of the roots is (-1)^n * c_0, where n is the degree. Wait, for quartic polynomial, product of roots is (-1)^4 * c_0 = c_0. Therefore, the constant term is equal to the product of the roots. Therefore, p(0) = product of roots. Wait, but in our case, the roots are 1, 2, 3, and a. Therefore, p(0) = 1 * 2 * 3 * a = 6a. Which matches our previous calculation.

Similarly, if we think about p(4), it's not directly related to Vieta's formula, but we can think of it as evaluating the polynomial at x = 4. However, since we already did that, and found that p(0) + p(4) = 24 regardless of a, that seems to be the case.

Alternatively, maybe there's a trick here. Let me think. Suppose we consider p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) + p(4) = [(-1)(-2)(-3)(-a)] + [(3)(2)(1)(4 - a)] = [6a] + [6(4 - a)] = 6a + 24 - 6a = 24. So that's another way to see it.

Therefore, regardless of the value of a, the answer is 24. Therefore, the answer is 24.

But let me check once more if there's any possible mistake. For example, if the polynomial had more roots or something else? But the problem says it's a degree 4 polynomial with three roots given, so the fourth is arbitrary. Therefore, the answer is always 24.

Alternatively, maybe the problem is designed so that even without knowing the fourth root, you can find p(0) + p(4). Which is exactly what we did.

Therefore, I think the answer is 24.

**Final Answer**
\boxed{24}

To solve the problem, we are given that $ p(x) $ is a **monic polynomial of degree 4**, and that **three of its roots are 1, 2, and 3**. We are asked to compute $ p(0) + p(4) $.

---

### Step 1: General Form of the Polynomial

Since $ p(x) $ is a monic polynomial of degree 4 and has roots 1, 2, 3, and an unknown fourth root $ a $, we can write:

$$
p(x) = (x - 1)(x - 2)(x - 3)(x - a)
$$

This is the factored form of the polynomial, and since it's monic, the leading coefficient is 1, which matches the given condition.

---

### Step 2: Compute $ p(0) $

Substitute $ x = 0 $ into the polynomial:

$$
p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a)
$$

Now compute the constants:

$$
(-1)(-2) = 2,\quad 2 \cdot (-3) = -6,\quad -6 \cdot (-a) = 6a
$$

So:

$$
p(0) = 6a
$$

---

### Step 3: Compute $ p(4) $

Substitute $ x = 4 $ into the polynomial:

$$
p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a)
$$

Compute the constants:

$$
3 \cdot 2 = 6,\quad 6 \cdot 1 = 6,\quad 6 \cdot (4 - a) = 24 - 6a
$$

So:

$$
p(4) = 24 - 6a
$$

---

### Step 4: Add $ p(0) + p(4) $

$$
p(0) + p(4) = 6a + (24 - 6a) = 24
$$

The terms involving $ a $ cancel out, and we are left with:

$$
\boxed{24}
$$

---

### Final Answer

$$
\boxed{24}
$$

[Assistant]
Explanation:
Okay, so I need to find p(0) + p(4) where p(x) is a monic polynomial of degree 4, and three of its roots are 1, 2, and 3. Let me start by recalling what a monic polynomial is. A monic polynomial is one where the leading coefficient (the coefficient of the highest power of x) is 1. Since it's a degree 4 polynomial, it has four roots (counting multiplicities). They told me three of the roots are 1, 2, and 3, so there must be a fourth root. Let me call that fourth root 'a'.

So, the polynomial can be written in factored form as p(x) = (x - 1)(x - 2)(x - 3)(x - a). Since it's monic, the leading coefficient is 1, which matches this form. Now, I need to figure out what 'a' is. Wait, but the problem doesn't give me any other information about the polynomial. Hmm. Maybe I don't need to find 'a' explicitly? Let me think.

The question asks for p(0) + p(4). Let me compute each term separately. Let's start with p(0). If I plug in x = 0 into the polynomial, I get p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a). Let me compute that. Multiplying the constants first: (-1)(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. So p(0) = 6a.

Now, let's compute p(4). Plugging in x = 4, we get p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a). Multiplying those constants: 3*2 is 6, 6*1 is 6, so p(4) = 6*(4 - a) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Wait, that's interesting. The 6a and -6a cancel each other out, leaving just 24. So regardless of the value of 'a', p(0) + p(4) is always 24? That seems surprising, but maybe that's correct. Let me check my steps again to make sure I didn't make a mistake.

First, p(x) is monic quartic with roots 1, 2, 3, and a. So factored form is correct. Then p(0) is product of (-1)(-2)(-3)(-a). Let me compute that again. (-1)*(-2) is 2, then 2*(-3) is -6, then -6*(-a) is 6a. Yes, that's correct. Then p(4) is (3)(2)(1)(4 - a) which is 6*(4 - a) = 24 - 6a. Then adding them together: 6a + 24 - 6a = 24. So yes, the answer is 24.

But wait, is there a possibility that the polynomial has a repeated root? The problem says "three of the roots are 1, 2, and 3". It doesn't specify if they are distinct or not. But since it's a degree 4 polynomial, and three roots are given, the fourth root could be any number, possibly one of the existing roots. However, even if there is a repeated root, the calculation for p(0) + p(4) would still hold because the fourth root is still 'a', whether it's equal to 1, 2, 3, or something else. For example, if a = 1, then p(x) would be (x - 1)^2(x - 2)(x - 3). But in that case, p(0) would be (-1)^2*(-2)*(-3) = 1*6 = 6, and p(4) would be (3)^2*(2)*(1) = 9*2 = 18. Then p(0) + p(4) = 6 + 18 = 24. Same result. Similarly, if a = 2, p(0) would be (-1)(-2)^2*(-3)(-2) = (-1)*(4)*(-3)*(-2). Wait, let me compute that again. Wait, if a = 2, then p(x) = (x - 1)(x - 2)^2(x - 3). Then p(0) = (-1)(-2)^2(-3) = (-1)*(4)*(-3) = 12. Then p(4) = (3)(2)^2(1) = 3*4*1 = 12. Then p(0) + p(4) = 12 + 12 = 24. Same result. Similarly, if a = 3, p(0) would be (-1)(-2)(-3)^2(-a) = (-1)(-2)(9)(-3) = (-1)(-2) is 2, 2*9 is 18, 18*(-3) is -54? Wait, hold on. Wait, if a = 3, then p(x) = (x - 1)(x - 2)(x - 3)^2. Then p(0) = (-1)(-2)(-3)^2 = (-1)(-2)(9) = 2*9 = 18. Then p(4) = (3)(2)(1)^2 = 3*2*1 = 6. Then p(0) + p(4) = 18 + 6 = 24. Still 24. So even if there are repeated roots, the answer is still 24. Therefore, regardless of the value of 'a', p(0) + p(4) is 24. Therefore, the answer is 24.

But let me think again. Is there a way that the problem could have more constraints? For example, maybe the polynomial is supposed to have distinct roots? But the problem didn't specify that. It just said three of the roots are 1, 2, 3. So the fourth root could be anything. Therefore, regardless of that, the answer is 24.

Alternatively, maybe there's another approach. Let me try expanding the polynomial. Let me write p(x) = (x - 1)(x - 2)(x - 3)(x - a). Let me first compute (x - 1)(x - 2)(x - 3). Let me compute that first. Let me compute (x - 1)(x - 2) first. That is x^2 - 3x + 2. Then multiply by (x - 3): (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Wait, let me do that step by step. Multiply x^2 by (x - 3): x^3 - 3x^2. Then -3x times (x - 3): -3x^2 + 9x. Then 2 times (x - 3): 2x - 6. Now add them all together: x^3 - 3x^2 - 3x^2 + 9x + 2x - 6. Combine like terms: x^3 - 6x^2 + 11x - 6. So (x - 1)(x - 2)(x - 3) = x^3 - 6x^2 + 11x - 6. Therefore, p(x) = (x^3 - 6x^2 + 11x - 6)(x - a). Let me expand this. Multiply each term by (x - a):

First, x^3*(x - a) = x^4 - a x^3.

Then -6x^2*(x - a) = -6x^3 + 6a x^2.

Then 11x*(x - a) = 11x^2 - 11a x.

Then -6*(x - a) = -6x + 6a.

Now combine all these terms:

x^4 - a x^3 - 6x^3 + 6a x^2 + 11x^2 - 11a x - 6x + 6a.

Combine like terms:

x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

So p(x) = x^4 + (-a - 6)x^3 + (6a + 11)x^2 + (-11a - 6)x + 6a.

Now, if I need to compute p(0) + p(4), let me compute p(0) first. p(0) is the constant term, which is 6a. Then p(4) is the value of the polynomial at x = 4. Let me compute p(4):

p(4) = (4)^4 + (-a - 6)(4)^3 + (6a + 11)(4)^2 + (-11a - 6)(4) + 6a.

Compute each term:

4^4 = 256.

(-a - 6)(4)^3 = (-a - 6)(64) = -64a - 384.

(6a + 11)(4)^2 = (6a + 11)(16) = 96a + 176.

(-11a - 6)(4) = -44a - 24.

And the last term is 6a.

So adding all these together:

256 + (-64a - 384) + (96a + 176) + (-44a - 24) + 6a.

Let me combine like terms step by step.

First, constants: 256 - 384 + 176 - 24.

Compute 256 - 384: that's -128.

Then -128 + 176: that's 48.

Then 48 - 24: that's 24.

Now, the terms with 'a':

-64a + 96a - 44a + 6a.

Compute -64a + 96a = 32a.

32a - 44a = -12a.

-12a + 6a = -6a.

So overall, p(4) = 24 - 6a.

Therefore, p(0) + p(4) = 6a + 24 - 6a = 24. Same result as before. So regardless of the value of 'a', the answer is 24. Therefore, the answer is 24.

But let me check once more with a specific example. Let me choose a specific value for 'a' and compute p(0) + p(4). Let's say a = 0. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 0) = x(x - 1)(x - 2)(x - 3). Then p(0) = 0, since there's an x term. Then p(4) = 4*(4 - 1)(4 - 2)(4 - 3) = 4*3*2*1 = 24. Therefore, p(0) + p(4) = 0 + 24 = 24. Correct.

Another example: let a = 4. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 4). Then p(0) = (-1)(-2)(-3)(-4) = 24. p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - 4) = 3*2*1*0 = 0. Therefore, p(0) + p(4) = 24 + 0 = 24. Correct again.

Another example: let a = 5. Then p(0) = (-1)(-2)(-3)(-5) = 30. p(4) = (3)(2)(1)(-1) = -6. Then p(0) + p(4) = 30 - 6 = 24. Correct.

So regardless of the value of 'a', the answer is always 24. Therefore, the answer is 24.

But wait, the problem says "three of the roots of p(x) are 1, 2, and 3". So maybe there's a possibility that the fourth root is something else? But even so, as shown, the answer is always 24. Therefore, the answer is 24.

Alternatively, maybe there's a different approach. Let me think. Since p(x) is a monic quartic with roots 1, 2, 3, and a. Then p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) is the product of the roots multiplied by (-1)^4, since for a monic polynomial, the constant term is (-1)^n times the product of the roots, where n is the degree. Wait, for a monic polynomial of degree 4, p(x) = x^4 + ... + (-1)^4 * product of roots. Wait, actually, the constant term is (-1)^4 times the product of the roots? Wait, no. Wait, the constant term is the product of the roots multiplied by (-1)^degree. Wait, let me recall Vieta's formula. For a monic polynomial p(x) = x^4 + c_3 x^3 + c_2 x^2 + c_1 x + c_0, the product of the roots is c_0 * (-1)^4 = c_0. Wait, Vieta's formula says that the product of the roots is (-1)^n * c_0, where n is the degree. Wait, for quartic polynomial, product of roots is (-1)^4 * c_0 = c_0. Therefore, the constant term is equal to the product of the roots. Therefore, p(0) = product of roots. Wait, but in our case, the roots are 1, 2, 3, and a. Therefore, p(0) = 1 * 2 * 3 * a = 6a. Which matches our previous calculation.

Similarly, if we think about p(4), it's not directly related to Vieta's formula, but we can think of it as evaluating the polynomial at x = 4. However, since we already did that, and found that p(0) + p(4) = 24 regardless of a, that seems to be the case.

Alternatively, maybe there's a trick here. Let me think. Suppose we consider p(x) = (x - 1)(x - 2)(x - 3)(x - a). Then p(0) + p(4) = [(-1)(-2)(-3)(-a)] + [(3)(2)(1)(4 - a)] = [6a] + [6(4 - a)] = 6a + 24 - 6a = 24. So that's another way to see it.

Therefore, regardless of the value of a, the answer is 24. Therefore, the answer is 24.

But let me check once more if there's any possible mistake. For example, if the polynomial had more roots or something else? But the problem says it's a degree 4 polynomial with three roots given, so the fourth is arbitrary. Therefore, the answer is always 24.

Alternatively, maybe the problem is designed so that even without knowing the fourth root, you can find p(0) + p(4). Which is exactly what we did.

Therefore, I think the answer is 24.

**Final Answer**
\boxed{24}

To solve the problem, we are given that $ p(x) $ is a **monic polynomial of degree 4**, and that **three of its roots are 1, 2, and 3**. We are asked to compute $ p(0) + p(4) $.

---

### Step 1: General Form of the Polynomial

Since $ p(x) $ is a monic polynomial of degree 4 and has roots 1, 2, 3, and an unknown fourth root $ a $, we can write:

$$
p(x) = (x - 1)(x - 2)(x - 3)(x - a)
$$

This is the factored form of the polynomial, and since it's monic, the leading coefficient is 1, which matches the given condition.

---

### Step 2: Compute $ p(0) $

Substitute $ x = 0 $ into the polynomial:

$$
p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - a) = (-1)(-2)(-3)(-a)
$$

Now compute the constants:

$$
(-1)(-2) = 2,\quad 2 \cdot (-3) = -6,\quad -6 \cdot (-a) = 6a
$$

So:

$$
p(0) = 6a
$$

---

### Step 3: Compute $ p(4) $

Substitute $ x = 4 $ into the polynomial:

$$
p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - a) = (3)(2)(1)(4 - a)
$$

Compute the constants:

$$
3 \cdot 2 = 6,\quad 6 \cdot 1 = 6,\quad 6 \cdot (4 - a) = 24 - 6a
$$

So:

$$
p(4) = 24 - 6a
$$

---

### Step 4: Add $ p(0) + p(4) $

$$
p(0) + p(4) = 6a + (24 - 6a) = 24
$$

The terms involving $ a $ cancel out, and we are left with:

$$
\boxed{24}
$$

---

### Final Answer

$$
\boxed{24}
$$

Solution:
Since three of the roots of $p(x)$ are 1, 2, and 3, we can write
\[p(x) = (x - 1)(x - 2)(x - 3)(x - r).\]Then
\begin{align*}
p(0) + p(4) &= (-1)(-2)(-3)(-r) + (3)(2)(1)(4 - r) \\
&= 6r + 24 - 6r = \boxed{24}.
\end{align*}


Explanation:
Okay, so I need to figure out how many daps equal 42 baps. The problem gives me two conversion rates: 4 daps = 7 yaps and 5 yaps = 3 baps. Hmm, let me think about how to approach this.

First, maybe I should find a way to convert baps to yaps and then yaps to daps? Or maybe convert daps to yaps and then yaps to baps? Let me see. Since the question is asking for daps in terms of baps, I need to relate daps and baps through yaps.

Let me write down the given equations:

1. 4 daps = 7 yaps
2. 5 yaps = 3 baps

So, if I can express baps in terms of yaps, and then yaps in terms of daps, I can link daps to baps. Let me try that.

Starting with the second equation: 5 yaps = 3 baps. If I want to find out how many yaps are in 1 bap, I can rearrange this equation. Dividing both sides by 3, I get:

(5/3) yaps = 1 bap

So, 1 bap is equal to 5/3 yaps. Alternatively, if I want to find how many baps are in 1 yap, I can divide both sides by 5:

1 yap = 3/5 baps

But maybe I need to go the other way. Since I need to convert baps to daps, perhaps I should first convert baps to yaps and then yaps to daps. Let me try that.

Starting with 42 baps. If I can convert baps to yaps first. From the second equation, 5 yaps = 3 baps. So, to find how many yaps are in 1 bap, as I had before, it's 5/3 yaps per bap. Therefore, for 42 baps, it would be:

42 baps * (5/3 yaps / 1 bap) = 42 * (5/3) yaps

Let me compute that. 42 divided by 3 is 14, so 14 * 5 = 70. So, 42 baps equal 70 yaps.

Okay, so 42 baps = 70 yaps. Now, I need to convert yaps to daps. From the first equation, 4 daps = 7 yaps. So, to find how many daps are in 1 yap, I can divide both sides by 7:

4/7 daps = 1 yap

Therefore, 1 yap is equal to 4/7 daps. Therefore, 70 yaps would be:

70 yaps * (4/7 daps / 1 yap) = 70 * (4/7) daps

Calculating that, 70 divided by 7 is 10, so 10 * 4 = 40. Therefore, 70 yaps equal 40 daps.

So, putting it all together: 42 baps = 70 yaps = 40 daps. Therefore, 42 baps equal 40 daps.

Wait, let me check my steps again to make sure I didn't make a mistake.

First, converting baps to yaps: 5 yaps = 3 baps. So, 1 bap = 5/3 yaps. Therefore, 42 baps = 42 * 5/3 yaps. 42 divided by 3 is 14, 14 * 5 is 70. That seems right.

Then converting yaps to daps: 4 daps = 7 yaps. So, 1 yap = 4/7 daps. Therefore, 70 yaps = 70 * 4/7 daps. 70 divided by 7 is 10, 10 * 4 is 40. That also seems right.

Alternatively, maybe I can do this using ratios or dimensional analysis. Let me try another approach to verify.

Let me set up the conversion factors.

We have 4 daps = 7 yaps. So, the conversion factor from daps to yaps is 7 yaps / 4 daps.

Similarly, 5 yaps = 3 baps, so the conversion factor from yaps to baps is 3 baps / 5 yaps.

But if I want to go from baps to daps, I need to reverse these.

Alternatively, maybe I can combine the two equations. Let me find a common term.

Let me express both equations in terms of daps and baps.

First equation: 4 daps = 7 yaps. Let me solve for yaps: yaps = (4/7) daps.

Second equation: 5 yaps = 3 baps. Let me solve for baps: baps = (5/3) yaps.

So, substituting the first equation into the second, we can write baps in terms of daps.

So, baps = (5/3) yaps = (5/3)*(4/7 daps) = (20/21) daps.

Wait, that would mean that baps = (20/21) daps? Wait, that seems like baps are less than daps? Wait, but if 4 daps = 7 yaps, and 5 yaps = 3 baps, then 4 daps = 7 yaps = (7/5)*3 baps = 21/5 baps. Therefore, 4 daps = 21/5 baps. Therefore, 1 dap = (21/5)/4 baps = 21/20 baps.

Wait, so 1 dap is 21/20 baps? Which is 1.05 baps. So, that would mean that 1 bap is 20/21 daps.

Wait, so if I want to find how many daps equal 42 baps, then since 1 bap = 20/21 daps, then 42 baps = 42*(20/21) daps.

Calculating that: 42 divided by 21 is 2, so 2*20 = 40. So, 42 baps = 40 daps.

So that's the same answer as before. So that seems to confirm that 40 is the correct answer.

Alternatively, maybe I can use proportions. Let me try that.

First, find how many daps are in 1 bap.

From the first equation: 4 daps = 7 yaps.

From the second equation: 5 yaps = 3 baps.

So, let me find how many daps are in 1 bap.

First, find how many yaps are in 1 bap: 5 yaps = 3 baps => 1 bap = 5/3 yaps.

Then, since 4 daps = 7 yaps, then 1 yap = 4/7 daps.

Therefore, 1 bap = 5/3 yaps * 4/7 daps per yap = (5/3)*(4/7) daps = 20/21 daps.

Therefore, 1 bap = 20/21 daps. Therefore, 42 baps = 42*(20/21) daps = 40 daps.

Same answer again.

Alternatively, maybe I can set up a proportion with all the units.

We need to find x such that x daps = 42 baps.

We can write the conversion factors as fractions.

From 4 daps = 7 yaps, so (4 daps)/(7 yaps) = 1.

From 5 yaps = 3 baps, so (5 yaps)/(3 baps) = 1.

Therefore, multiplying these two equations together:

(4 daps)/(7 yaps) * (5 yaps)/(3 baps) = 1 * 1

Simplify: (4*5 daps yaps) / (7*3 yaps baps) = 1

The yaps cancel out: (20 daps) / (21 baps) = 1

Therefore, 20 daps = 21 baps

Therefore, 1 bap = 20/21 daps

Therefore, 42 baps = 42*(20/21) daps = 40 daps

Same result again.

So, all methods lead to 40 daps.

Alternatively, maybe I can use unit cancellation.

Starting with 42 baps, and converting to daps.

We have 42 baps * (5 yaps / 3 baps) * (4 daps / 7 yaps)

Let me compute that.

First, the baps unit cancels with the baps in the denominator.

So, 42 * (5/3) yaps.

Then, the yaps unit cancels with the yaps in the denominator.

So, 42 * (5/3) * (4/7) daps.

Calculating the numerical value:

First, 42 * (5/3) = 42*(5)/3 = 14*5 = 70

Then, 70 * (4/7) = 70*(4)/7 = 10*4 = 40

So, again, 40 daps.

Therefore, regardless of the method, the answer is 40.

I think that's solid. All different approaches confirm that 42 baps equal 40 daps.

But just to make sure, let me check with actual numbers.

Suppose I have 4 daps = 7 yaps. Let me find how many yaps are in 4 daps: 7 yaps.

Then, 5 yaps = 3 baps. So, 7 yaps would be equal to (7/5)*3 baps = 21/5 baps = 4.2 baps.

Therefore, 4 daps = 4.2 baps.

Therefore, 1 dap = 4.2 / 4 = 1.05 baps.

Therefore, 1 bap = 1 / 1.05 daps ≈ 0.95238 daps.

But 20/21 is approximately 0.95238, which matches.

Therefore, 1 bap ≈ 0.95238 daps.

Therefore, 42 baps ≈ 42 * 0.95238 ≈ 40 daps.

Yes, that's correct.

Alternatively, if 4 daps = 4.2 baps, then how many daps equal 42 baps?

Set up a proportion:

4 daps / 4.2 baps = x daps / 42 baps

Cross-multiplying: 4.2x = 4 * 42

Therefore, x = (4 * 42)/4.2

Calculate that: 4*42 = 168

168 / 4.2 = 40

Yes, same answer.

Therefore, all methods confirm that 42 baps equal 40 daps.

I think that's thorough enough. I can't see any mistakes in my reasoning. All different approaches lead to the same answer, so I'm confident that 40 is the correct number of daps equal to 42 baps.

**Final Answer**
The number of daps equal to 42 baps is \boxed{40}.

To determine how many **daps** are equal to **42 baps**, we are given two conversion rates:

1. **4 daps = 7 yaps**
2. **5 yaps = 3 baps**

Our goal is to convert **baps to daps**, so we need to find a way to relate **baps** to **daps** through **yaps**.

---

### Step 1: Convert baps to yaps

From the second equation:

$$
5 \text{ yaps} = 3 \text{ baps} \Rightarrow 1 \text{ bap} = \frac{5}{3} \text{ yaps}
$$

So, for **42 baps**:

$$
42 \text{ baps} \times \frac{5}{3} \text{ yaps/bap} = \frac{42 \times 5}{3} = \frac{210}{3} = 70 \text{ yaps}
$$

---

### Step 2: Convert yaps to daps

From the first equation:

$$
4 \text{ daps} = 7 \text{ yaps} \Rightarrow 1 \text{ yap} = \frac{4}{7} \text{ daps}
$$

So, for **70 yaps**:

$$
70 \text{ yaps} \times \frac{4}{7} \text{ daps/yap} = \frac{70 \times 4}{7} = \frac{280}{7} = 40 \text{ daps}
$$

---

### Final Answer

$$
\boxed{40}
$$

So, **42 baps** are equal to **40 daps**.

[Assistant]
Explanation:
Okay, so I need to figure out how many daps equal 42 baps. The problem gives me two conversion rates: 4 daps = 7 yaps and 5 yaps = 3 baps. Hmm, let me think about how to approach this.

First, maybe I should find a way to convert baps to yaps and then yaps to daps? Or maybe convert daps to yaps and then yaps to baps? Let me see. Since the question is asking for daps in terms of baps, I need to relate daps and baps through yaps.

Let me write down the given equations:

1. 4 daps = 7 yaps
2. 5 yaps = 3 baps

So, if I can express baps in terms of yaps, and then yaps in terms of daps, I can link daps to baps. Let me try that.

Starting with the second equation: 5 yaps = 3 baps. If I want to find out how many yaps are in 1 bap, I can rearrange this equation. Dividing both sides by 3, I get:

(5/3) yaps = 1 bap

So, 1 bap is equal to 5/3 yaps. Alternatively, if I want to find how many baps are in 1 yap, I can divide both sides by 5:

1 yap = 3/5 baps

But maybe I need to go the other way. Since I need to convert baps to daps, perhaps I should first convert baps to yaps and then yaps to daps. Let me try that.

Starting with 42 baps. If I can convert baps to yaps first. From the second equation, 5 yaps = 3 baps. So, to find how many yaps are in 1 bap, as I had before, it's 5/3 yaps per bap. Therefore, for 42 baps, it would be:

42 baps * (5/3 yaps / 1 bap) = 42 * (5/3) yaps

Let me compute that. 42 divided by 3 is 14, so 14 * 5 = 70. So, 42 baps equal 70 yaps.

Okay, so 42 baps = 70 yaps. Now, I need to convert yaps to daps. From the first equation, 4 daps = 7 yaps. So, to find how many daps are in 1 yap, I can divide both sides by 7:

4/7 daps = 1 yap

Therefore, 1 yap is equal to 4/7 daps. Therefore, 70 yaps would be:

70 yaps * (4/7 daps / 1 yap) = 70 * (4/7) daps

Calculating that, 70 divided by 7 is 10, so 10 * 4 = 40. Therefore, 70 yaps equal 40 daps.

So, putting it all together: 42 baps = 70 yaps = 40 daps. Therefore, 42 baps equal 40 daps.

Wait, let me check my steps again to make sure I didn't make a mistake.

First, converting baps to yaps: 5 yaps = 3 baps. So, 1 bap = 5/3 yaps. Therefore, 42 baps = 42 * 5/3 yaps. 42 divided by 3 is 14, 14 * 5 is 70. That seems right.

Then converting yaps to daps: 4 daps = 7 yaps. So, 1 yap = 4/7 daps. Therefore, 70 yaps = 70 * 4/7 daps. 70 divided by 7 is 10, 10 * 4 is 40. That also seems right.

Alternatively, maybe I can do this using ratios or dimensional analysis. Let me try another approach to verify.

Let me set up the conversion factors.

We have 4 daps = 7 yaps. So, the conversion factor from daps to yaps is 7 yaps / 4 daps.

Similarly, 5 yaps = 3 baps, so the conversion factor from yaps to baps is 3 baps / 5 yaps.

But if I want to go from baps to daps, I need to reverse these.

Alternatively, maybe I can combine the two equations. Let me find a common term.

Let me express both equations in terms of daps and baps.

First equation: 4 daps = 7 yaps. Let me solve for yaps: yaps = (4/7) daps.

Second equation: 5 yaps = 3 baps. Let me solve for baps: baps = (5/3) yaps.

So, substituting the first equation into the second, we can write baps in terms of daps.

So, baps = (5/3) yaps = (5/3)*(4/7 daps) = (20/21) daps.

Wait, that would mean that baps = (20/21) daps? Wait, that seems like baps are less than daps? Wait, but if 4 daps = 7 yaps, and 5 yaps = 3 baps, then 4 daps = 7 yaps = (7/5)*3 baps = 21/5 baps. Therefore, 4 daps = 21/5 baps. Therefore, 1 dap = (21/5)/4 baps = 21/20 baps.

Wait, so 1 dap is 21/20 baps? Which is 1.05 baps. So, that would mean that 1 bap is 20/21 daps.

Wait, so if I want to find how many daps equal 42 baps, then since 1 bap = 20/21 daps, then 42 baps = 42*(20/21) daps.

Calculating that: 42 divided by 21 is 2, so 2*20 = 40. So, 42 baps = 40 daps.

So that's the same answer as before. So that seems to confirm that 40 is the correct answer.

Alternatively, maybe I can use proportions. Let me try that.

First, find how many daps are in 1 bap.

From the first equation: 4 daps = 7 yaps.

From the second equation: 5 yaps = 3 baps.

So, let me find how many daps are in 1 bap.

First, find how many yaps are in 1 bap: 5 yaps = 3 baps => 1 bap = 5/3 yaps.

Then, since 4 daps = 7 yaps, then 1 yap = 4/7 daps.

Therefore, 1 bap = 5/3 yaps * 4/7 daps per yap = (5/3)*(4/7) daps = 20/21 daps.

Therefore, 1 bap = 20/21 daps. Therefore, 42 baps = 42*(20/21) daps = 40 daps.

Same answer again.

Alternatively, maybe I can set up a proportion with all the units.

We need to find x such that x daps = 42 baps.

We can write the conversion factors as fractions.

From 4 daps = 7 yaps, so (4 daps)/(7 yaps) = 1.

From 5 yaps = 3 baps, so (5 yaps)/(3 baps) = 1.

Therefore, multiplying these two equations together:

(4 daps)/(7 yaps) * (5 yaps)/(3 baps) = 1 * 1

Simplify: (4*5 daps yaps) / (7*3 yaps baps) = 1

The yaps cancel out: (20 daps) / (21 baps) = 1

Therefore, 20 daps = 21 baps

Therefore, 1 bap = 20/21 daps

Therefore, 42 baps = 42*(20/21) daps = 40 daps

Same result again.

So, all methods lead to 40 daps.

Alternatively, maybe I can use unit cancellation.

Starting with 42 baps, and converting to daps.

We have 42 baps * (5 yaps / 3 baps) * (4 daps / 7 yaps)

Let me compute that.

First, the baps unit cancels with the baps in the denominator.

So, 42 * (5/3) yaps.

Then, the yaps unit cancels with the yaps in the denominator.

So, 42 * (5/3) * (4/7) daps.

Calculating the numerical value:

First, 42 * (5/3) = 42*(5)/3 = 14*5 = 70

Then, 70 * (4/7) = 70*(4)/7 = 10*4 = 40

So, again, 40 daps.

Therefore, regardless of the method, the answer is 40.

I think that's solid. All different approaches confirm that 42 baps equal 40 daps.

But just to make sure, let me check with actual numbers.

Suppose I have 4 daps = 7 yaps. Let me find how many yaps are in 4 daps: 7 yaps.

Then, 5 yaps = 3 baps. So, 7 yaps would be equal to (7/5)*3 baps = 21/5 baps = 4.2 baps.

Therefore, 4 daps = 4.2 baps.

Therefore, 1 dap = 4.2 / 4 = 1.05 baps.

Therefore, 1 bap = 1 / 1.05 daps ≈ 0.95238 daps.

But 20/21 is approximately 0.95238, which matches.

Therefore, 1 bap ≈ 0.95238 daps.

Therefore, 42 baps ≈ 42 * 0.95238 ≈ 40 daps.

Yes, that's correct.

Alternatively, if 4 daps = 4.2 baps, then how many daps equal 42 baps?

Set up a proportion:

4 daps / 4.2 baps = x daps / 42 baps

Cross-multiplying: 4.2x = 4 * 42

Therefore, x = (4 * 42)/4.2

Calculate that: 4*42 = 168

168 / 4.2 = 40

Yes, same answer.

Therefore, all methods confirm that 42 baps equal 40 daps.

I think that's thorough enough. I can't see any mistakes in my reasoning. All different approaches lead to the same answer, so I'm confident that 40 is the correct number of daps equal to 42 baps.

**Final Answer**
The number of daps equal to 42 baps is \boxed{40}.

To determine how many **daps** are equal to **42 baps**, we are given two conversion rates:

1. **4 daps = 7 yaps**
2. **5 yaps = 3 baps**

Our goal is to convert **baps to daps**, so we need to find a way to relate **baps** to **daps** through **yaps**.

---

### Step 1: Convert baps to yaps

From the second equation:

$$
5 \text{ yaps} = 3 \text{ baps} \Rightarrow 1 \text{ bap} = \frac{5}{3} \text{ yaps}
$$

So, for **42 baps**:

$$
42 \text{ baps} \times \frac{5}{3} \text{ yaps/bap} = \frac{42 \times 5}{3} = \frac{210}{3} = 70 \text{ yaps}
$$

---

### Step 2: Convert yaps to daps

From the first equation:

$$
4 \text{ daps} = 7 \text{ yaps} \Rightarrow 1 \text{ yap} = \frac{4}{7} \text{ daps}
$$

So, for **70 yaps**:

$$
70 \text{ yaps} \times \frac{4}{7} \text{ daps/yap} = \frac{70 \times 4}{7} = \frac{280}{7} = 40 \text{ daps}
$$

---

### Final Answer

$$
\boxed{40}
$$

So, **42 baps** are equal to **40 daps**.

Solution: Multiply $5\text{ yaps}=3\text{ baps}$ by 14 to find that 70 yaps are equal to 42 baps.  Then multiply $4\text{ daps}=7\text{ yaps}$ by 10 to find that $\boxed{40}$ daps are equal to 70 yaps.


Explanation:
Okay, so there's this problem about a group of N students, and N is less than 50. The teacher tried grouping them in different ways, and we have some conditions. Let me try to parse this step by step.

First, when the teacher puts them in groups of 8, the last group has 5 students. That means if you divide N by 8, the remainder is 5. So mathematically, that would be N ≡ 5 mod 8. Right? Because when you divide by 8, the remainder is 5. So N = 8k + 5 for some integer k.

Then, if the teacher puts them in groups of 6, the last group has 3 students. So similarly, N ≡ 3 mod 6. So N = 6m + 3 for some integer m.

So we need to find all numbers N less than 50 that satisfy both congruences: N ≡ 5 mod 8 and N ≡ 3 mod 6. Then sum all those possible Ns.

Alright, so how do I solve this system of congruences? Maybe using the Chinese Remainder Theorem? Or maybe just listing numbers that satisfy one condition and checking the other?

Since N is less than 50, maybe listing numbers that satisfy N ≡ 5 mod 8 first. Let me try that.

Numbers congruent to 5 mod 8 less than 50:

Start with 5, then 5 + 8 = 13, 13 + 8 = 21, 21 + 8 = 29, 29 + 8 = 37, 37 + 8 = 45, 45 + 8 = 53. But 53 is over 50, so stop there. So possible Ns are 5, 13, 21, 29, 37, 45.

Now check which of these numbers also satisfy N ≡ 3 mod 6.

Let me check each one:

First, N = 5. Let's divide 5 by 6. The remainder is 5. But we need remainder 3. So 5 is out.

Next, N = 13. 13 divided by 6 is 2 with remainder 1. So remainder 1. Not 3. So nope.

Next, N = 21. 21 divided by 6 is 3 with remainder 3. Oh! That works. So 21 is a solution.

Next, N = 29. 29 divided by 6: 6*4=24, 29-24=5. Remainder 5. Not 3. So no.

Next, N = 37. 37 divided by 6: 6*6=36, 37-36=1. Remainder 1. Not 3.

Next, N = 45. 45 divided by 6: 6*7=42, 45-42=3. Remainder 3. So that works. So 45 is another solution.

So the numbers are 21 and 45. Wait, so sum of all possible values is 21 + 45 = 66?

But let me check if there are more numbers. Wait, the numbers congruent to 5 mod 8 less than 50 are 5,13,21,29,37,45. We checked all of them. Only 21 and 45 satisfy the second condition. So sum is 66. Is that the answer?

Wait, but let me make sure I didn't miss any numbers. Maybe there's another way to approach this?

Alternatively, solving the congruences:

We have N ≡ 5 mod 8 and N ≡ 3 mod 6.

Let me write N = 8k + 5. Substitute into the second equation:

8k + 5 ≡ 3 mod 6.

So 8k + 5 ≡ 3 mod 6. Let's simplify 8k mod 6. Since 8 ≡ 2 mod 6, so 8k ≡ 2k mod 6. Therefore:

2k + 5 ≡ 3 mod 6.

Subtract 5 from both sides:

2k ≡ 3 - 5 mod 6 => 2k ≡ -2 mod 6 => which is same as 2k ≡ 4 mod 6 (since -2 + 6 = 4).

So 2k ≡ 4 mod 6. Divide both sides by 2? Wait, but in modular arithmetic, dividing by 2 is only possible if 2 and 6 are coprime? Wait, 2 and 6 have a common factor of 2. So we can divide the congruence by 2, but we have to divide the modulus by gcd(2,6)=2 as well. So:

Dividing both sides by 2: k ≡ 2 mod 3. Because 2k ≡ 4 mod 6 => k ≡ 2 mod 3.

So k = 3m + 2 for some integer m.

Therefore, N = 8k + 5 = 8*(3m + 2) + 5 = 24m + 16 + 5 = 24m + 21.

Therefore, N ≡ 21 mod 24. So the solutions are numbers congruent to 21 mod 24. So less than 50, so:

First solution is 21, then 21 + 24 = 45, then 45 + 24 = 69, which is over 50. So only 21 and 45. Therefore, sum is 21 + 45 = 66. So same as before. So answer is 66?

Wait, but let me check if there are other solutions. For example, if m is negative? But since N has to be positive, m can't be negative. For m=0: N=21. m=1: 45. m=2: 69, which is over 50. So yes, only 21 and 45. So sum is 66. So answer is 66?

But let me check again with the original problem. Let me check N=21:

Divided by 8: 21 /8 is 2 groups of 8, which is 16, remainder 5. Correct. Divided by 6: 21 /6 is 3 groups of 6, which is 18, remainder 3. Correct. So that works.

N=45: Divided by 8: 45 /8 is 5*8=40, remainder 5. Correct. Divided by 6: 45 /6 is 7*6=42, remainder 3. Correct. So both work.

So the possible values are 21 and 45. Sum is 66. Therefore, the answer is 66. Wait, but let me check if there are other numbers. For example, maybe numbers that satisfy both congruences but not in the list? But according to the equations, the solutions are N ≡21 mod 24. So numbers 21, 45, 69,... So under 50, only 21 and 45. So yes, sum is 66.

But let me check if there's another way someone might approach this problem and maybe get a different answer? For example, maybe miscalculating the congruence?

Wait, let me check solving the congruence again. So N ≡5 mod8 and N≡3 mod6.

Another way: Let me list numbers congruent to 3 mod6 less than 50 and check which are congruent to 5 mod8.

Numbers congruent to 3 mod6: 3, 9, 15, 21, 27, 33, 39, 45, 51,... So up to 45.

Now check which of these numbers are congruent to 5 mod8.

Check 3: 3 mod8 is 3. Not 5.

9: 9 mod8 is 1. Not 5.

15: 15 mod8 is 7. Not 5.

21: 21 mod8 is 5. Yes! So 21 is there.

27: 27 mod8 is 3 (since 3*8=24, 27-24=3). Not 5.

33: 33 mod8: 33-32=1. 1. Not 5.

39: 39-32=7. 7 mod8. Not 5.

45: 45-40=5. 5 mod8. Yes! So 45. So same numbers: 21 and 45. So same result. So sum is 66. So answer is 66.

Therefore, I think that's correct. So the sum of all possible values of N is 21 + 45 = 66.

But let me check once more if there's a possibility of another number. For example, maybe N= 21 - 24= -3? No, negative numbers don't make sense here. So no. So only 21 and 45. So answer is 66. So I think that's correct.

**Final Answer**
The sum of all possible values of $ N $ is \boxed{66}.

To solve the problem, we are given two conditions about the number of students $ N $, where $ N < 50 $:

1. When the students are grouped into groups of 8, the last group has 5 students.
   This implies:
   $$
   N \equiv 5 \pmod{8}
   $$

2. When the students are grouped into groups of 6, the last group has 3 students.
   This implies:
   $$
   N \equiv 3 \pmod{6}
   $$

We are to find all such values of $ N $ that satisfy both congruences and are less than 50, and then compute the sum of these values.

---

### Step 1: Solve the System of Congruences

We are solving the system:
$$
\begin{cases}
N \equiv 5 \pmod{8} \\
N \equiv 3 \pmod{6}
\end{cases}
$$

Let’s express $ N $ in terms of the first congruence:
$$
N = 8k + 5
$$
Substitute into the second congruence:
$$
8k + 5 \equiv 3 \pmod{6}
$$
Simplify:
$$
8k \equiv -2 \pmod{6} \Rightarrow 8k \equiv 4 \pmod{6}
$$
Since $ 8 \equiv 2 \pmod{6} $, this becomes:
$$
2k \equiv 4 \pmod{6}
$$
Divide both sides by 2 (since 2 and 6 are not coprime, we must divide the modulus as well):
$$
k \equiv 2 \pmod{3}
$$
So $ k = 3m + 2 $ for some integer $ m $, and substituting back:
$$
N = 8k + 5 = 8(3m + 2) + 5 = 24m + 21
$$
Thus, the general solution is:
$$
N \equiv 21 \pmod{24}
$$

---

### Step 2: Find All Valid $ N < 50 $

We now find all values of $ N $ less than 50 that satisfy $ N \equiv 21 \pmod{24} $:

- $ N = 21 $
- $ N = 21 + 24 = 45 $
- $ N = 45 + 24 = 69 $ (too large)

So the valid values are:
$$
N = 21, 45
$$

---

### Step 3: Compute the Sum

$$
21 + 45 = \boxed{66}
$$

---

### Final Answer

$$
\boxed{66}
$$

[Assistant]
Explanation:
Okay, so there's this problem about a group of N students, and N is less than 50. The teacher tried grouping them in different ways, and we have some conditions. Let me try to parse this step by step.

First, when the teacher puts them in groups of 8, the last group has 5 students. That means if you divide N by 8, the remainder is 5. So mathematically, that would be N ≡ 5 mod 8. Right? Because when you divide by 8, the remainder is 5. So N = 8k + 5 for some integer k.

Then, if the teacher puts them in groups of 6, the last group has 3 students. So similarly, N ≡ 3 mod 6. So N = 6m + 3 for some integer m.

So we need to find all numbers N less than 50 that satisfy both congruences: N ≡ 5 mod 8 and N ≡ 3 mod 6. Then sum all those possible Ns.

Alright, so how do I solve this system of congruences? Maybe using the Chinese Remainder Theorem? Or maybe just listing numbers that satisfy one condition and checking the other?

Since N is less than 50, maybe listing numbers that satisfy N ≡ 5 mod 8 first. Let me try that.

Numbers congruent to 5 mod 8 less than 50:

Start with 5, then 5 + 8 = 13, 13 + 8 = 21, 21 + 8 = 29, 29 + 8 = 37, 37 + 8 = 45, 45 + 8 = 53. But 53 is over 50, so stop there. So possible Ns are 5, 13, 21, 29, 37, 45.

Now check which of these numbers also satisfy N ≡ 3 mod 6.

Let me check each one:

First, N = 5. Let's divide 5 by 6. The remainder is 5. But we need remainder 3. So 5 is out.

Next, N = 13. 13 divided by 6 is 2 with remainder 1. So remainder 1. Not 3. So nope.

Next, N = 21. 21 divided by 6 is 3 with remainder 3. Oh! That works. So 21 is a solution.

Next, N = 29. 29 divided by 6: 6*4=24, 29-24=5. Remainder 5. Not 3. So no.

Next, N = 37. 37 divided by 6: 6*6=36, 37-36=1. Remainder 1. Not 3.

Next, N = 45. 45 divided by 6: 6*7=42, 45-42=3. Remainder 3. So that works. So 45 is another solution.

So the numbers are 21 and 45. Wait, so sum of all possible values is 21 + 45 = 66?

But let me check if there are more numbers. Wait, the numbers congruent to 5 mod 8 less than 50 are 5,13,21,29,37,45. We checked all of them. Only 21 and 45 satisfy the second condition. So sum is 66. Is that the answer?

Wait, but let me make sure I didn't miss any numbers. Maybe there's another way to approach this?

Alternatively, solving the congruences:

We have N ≡ 5 mod 8 and N ≡ 3 mod 6.

Let me write N = 8k + 5. Substitute into the second equation:

8k + 5 ≡ 3 mod 6.

So 8k + 5 ≡ 3 mod 6. Let's simplify 8k mod 6. Since 8 ≡ 2 mod 6, so 8k ≡ 2k mod 6. Therefore:

2k + 5 ≡ 3 mod 6.

Subtract 5 from both sides:

2k ≡ 3 - 5 mod 6 => 2k ≡ -2 mod 6 => which is same as 2k ≡ 4 mod 6 (since -2 + 6 = 4).

So 2k ≡ 4 mod 6. Divide both sides by 2? Wait, but in modular arithmetic, dividing by 2 is only possible if 2 and 6 are coprime? Wait, 2 and 6 have a common factor of 2. So we can divide the congruence by 2, but we have to divide the modulus by gcd(2,6)=2 as well. So:

Dividing both sides by 2: k ≡ 2 mod 3. Because 2k ≡ 4 mod 6 => k ≡ 2 mod 3.

So k = 3m + 2 for some integer m.

Therefore, N = 8k + 5 = 8*(3m + 2) + 5 = 24m + 16 + 5 = 24m + 21.

Therefore, N ≡ 21 mod 24. So the solutions are numbers congruent to 21 mod 24. So less than 50, so:

First solution is 21, then 21 + 24 = 45, then 45 + 24 = 69, which is over 50. So only 21 and 45. Therefore, sum is 21 + 45 = 66. So same as before. So answer is 66?

Wait, but let me check if there are other solutions. For example, if m is negative? But since N has to be positive, m can't be negative. For m=0: N=21. m=1: 45. m=2: 69, which is over 50. So yes, only 21 and 45. So sum is 66. So answer is 66?

But let me check again with the original problem. Let me check N=21:

Divided by 8: 21 /8 is 2 groups of 8, which is 16, remainder 5. Correct. Divided by 6: 21 /6 is 3 groups of 6, which is 18, remainder 3. Correct. So that works.

N=45: Divided by 8: 45 /8 is 5*8=40, remainder 5. Correct. Divided by 6: 45 /6 is 7*6=42, remainder 3. Correct. So both work.

So the possible values are 21 and 45. Sum is 66. Therefore, the answer is 66. Wait, but let me check if there are other numbers. For example, maybe numbers that satisfy both congruences but not in the list? But according to the equations, the solutions are N ≡21 mod 24. So numbers 21, 45, 69,... So under 50, only 21 and 45. So yes, sum is 66.

But let me check if there's another way someone might approach this problem and maybe get a different answer? For example, maybe miscalculating the congruence?

Wait, let me check solving the congruence again. So N ≡5 mod8 and N≡3 mod6.

Another way: Let me list numbers congruent to 3 mod6 less than 50 and check which are congruent to 5 mod8.

Numbers congruent to 3 mod6: 3, 9, 15, 21, 27, 33, 39, 45, 51,... So up to 45.

Now check which of these numbers are congruent to 5 mod8.

Check 3: 3 mod8 is 3. Not 5.

9: 9 mod8 is 1. Not 5.

15: 15 mod8 is 7. Not 5.

21: 21 mod8 is 5. Yes! So 21 is there.

27: 27 mod8 is 3 (since 3*8=24, 27-24=3). Not 5.

33: 33 mod8: 33-32=1. 1. Not 5.

39: 39-32=7. 7 mod8. Not 5.

45: 45-40=5. 5 mod8. Yes! So 45. So same numbers: 21 and 45. So same result. So sum is 66. So answer is 66.

Therefore, I think that's correct. So the sum of all possible values of N is 21 + 45 = 66.

But let me check once more if there's a possibility of another number. For example, maybe N= 21 - 24= -3? No, negative numbers don't make sense here. So no. So only 21 and 45. So answer is 66. So I think that's correct.

**Final Answer**
The sum of all possible values of $ N $ is \boxed{66}.

To solve the problem, we are given two conditions about the number of students $ N $, where $ N < 50 $:

1. When the students are grouped into groups of 8, the last group has 5 students.
   This implies:
   $$
   N \equiv 5 \pmod{8}
   $$

2. When the students are grouped into groups of 6, the last group has 3 students.
   This implies:
   $$
   N \equiv 3 \pmod{6}
   $$

We are to find all such values of $ N $ that satisfy both congruences and are less than 50, and then compute the sum of these values.

---

### Step 1: Solve the System of Congruences

We are solving the system:
$$
\begin{cases}
N \equiv 5 \pmod{8} \\
N \equiv 3 \pmod{6}
\end{cases}
$$

Let’s express $ N $ in terms of the first congruence:
$$
N = 8k + 5
$$
Substitute into the second congruence:
$$
8k + 5 \equiv 3 \pmod{6}
$$
Simplify:
$$
8k \equiv -2 \pmod{6} \Rightarrow 8k \equiv 4 \pmod{6}
$$
Since $ 8 \equiv 2 \pmod{6} $, this becomes:
$$
2k \equiv 4 \pmod{6}
$$
Divide both sides by 2 (since 2 and 6 are not coprime, we must divide the modulus as well):
$$
k \equiv 2 \pmod{3}
$$
So $ k = 3m + 2 $ for some integer $ m $, and substituting back:
$$
N = 8k + 5 = 8(3m + 2) + 5 = 24m + 21
$$
Thus, the general solution is:
$$
N \equiv 21 \pmod{24}
$$

---

### Step 2: Find All Valid $ N < 50 $

We now find all values of $ N $ less than 50 that satisfy $ N \equiv 21 \pmod{24} $:

- $ N = 21 $
- $ N = 21 + 24 = 45 $
- $ N = 45 + 24 = 69 $ (too large)

So the valid values are:
$$
N = 21, 45
$$

---

### Step 3: Compute the Sum

$$
21 + 45 = \boxed{66}
$$

---

### Final Answer

$$
\boxed{66}
$$

Solution: We are given that $N\equiv 5\pmod{8}$ and $N\equiv 3\pmod{6}$.  We begin checking numbers which are 5 more than a multiple of 8, and we find that 5 and 13 are not 3 more than a multiple of 6, but 21 is 3 more than a multiple of 6. Thus 21 is one possible value of $N$. By the Chinese Remainder Theorem, the integers $x$ satisfying $x\equiv 5\pmod{8}$ and $x\equiv 3\pmod{6}$ are those of the form $x=21+\text{lcm}(6,8)k = 21 + 24 k$, where $k$ is an integer. Thus the 2 solutions less than $50$ are 21 and $21+24(1) = 45$, and their sum is $21+45=\boxed{66}$.