Explanation:
Okay, let me try to figure out how to solve this problem. The question is asking me to write the expression √2 + 1/√2 + √3 + 1/√3 in the form (a√2 + b√3)/c, where a, b, and c are positive integers, and c is as small as possible. Then, I need to find a + b + c. Alright, let me start by breaking down the problem step by step.

First, let me write down the original expression:

√2 + 1/√2 + √3 + 1/√3.

Hmm, I need to combine these terms into a single fraction with the form (a√2 + b√3)/c. To do that, I probably need to simplify each term first, maybe rationalize the denominators where necessary. Let me recall how to rationalize denominators. For example, 1/√2 can be rationalized by multiplying numerator and denominator by √2, right?

Let me try that. Let me handle each term one by one.

Starting with √2. Well, that's already simplified. Then 1/√2. If I rationalize that, it becomes √2/2. Similarly, for 1/√3, rationalizing gives √3/3.

So let me rewrite the entire expression with these rationalized forms:

√2 + √2/2 + √3 + √3/3.

Okay, now I can combine like terms. Let me group the terms with √2 and the terms with √3.

For the √2 terms: √2 + √2/2. Let me compute that. To add these, I need a common denominator. The denominators here are 1 and 2, so common denominator is 2. So √2 is equal to 2√2/2. Therefore, adding 2√2/2 + √2/2 gives (2√2 + √2)/2 = 3√2/2.

Similarly, for the √3 terms: √3 + √3/3. Again, common denominator for 1 and 3 is 3. So √3 is 3√3/3. Therefore, adding 3√3/3 + √3/3 gives (3√3 + √3)/3 = 4√3/3.

So now the entire expression simplifies to 3√2/2 + 4√3/3. Now, the problem wants this in the form (a√2 + b√3)/c. So right now, I have two separate fractions: 3√2/2 and 4√3/3. To combine them into a single fraction, I need to find a common denominator. The denominators are 2 and 3, so the least common denominator is 6. Let me convert both fractions to have denominator 6.

First term: 3√2/2. Multiply numerator and denominator by 3: (3√2 * 3)/(2 * 3) = 9√2/6.

Second term: 4√3/3. Multiply numerator and denominator by 2: (4√3 * 2)/(3 * 2) = 8√3/6.

Now, adding these two fractions together: 9√2/6 + 8√3/6 = (9√2 + 8√3)/6.

So the expression simplifies to (9√2 + 8√3)/6. Now, check if this is in the desired form. The problem says that a, b, c should be positive integers and c is as small as possible. Here, a is 9, b is 8, and c is 6. Let me check if this fraction can be simplified further. For that, I need to see if 9, 8, and 6 have a common factor.

Looking at 9, 8, and 6. The factors:

- 9 factors: 1, 3, 9

- 8 factors: 1, 2, 4, 8

- 6 factors: 1, 2, 3, 6

The greatest common divisor (GCD) of all three numbers 9, 8, and 6. The GCD of 9 and 8 is 1, since they are co-prime. Therefore, the GCD of all three numbers is 1. Therefore, the fraction (9√2 + 8√3)/6 cannot be simplified further. Therefore, c is 6, which is already as small as possible.

Therefore, a = 9, b = 8, c = 6. Therefore, a + b + c = 9 + 8 + 6 = 23. Is that correct? Let me double-check my steps to make sure I didn't make any mistakes.

Starting again: Original expression √2 + 1/√2 + √3 + 1/√3. Rationalizing denominators:

√2 + √2/2 + √3 + √3/3. Combining √2 terms: √2 + √2/2 = (2√2 + √2)/2 = 3√2/2. Similarly for √3 terms: √3 + √3/3 = (3√3 + √3)/3 = 4√3/3. Then adding those two fractions: 3√2/2 + 4√3/3. Common denominator 6: 9√2/6 + 8√3/6 = (9√2 + 8√3)/6. So that seems correct. GCD of numerator coefficients (9,8,6) is 1, so c is indeed 6. So a + b + c is 9 + 8 + 6 = 23. Hmm.

Wait, but let me check if there's another way to combine the original terms that might lead to a different form with smaller c. Maybe I made a mistake in assuming that combining the terms as I did is the only way? Let me think.

Alternatively, perhaps I can combine all the terms over a common denominator from the start? Let me try that approach.

Original expression:

√2 + 1/√2 + √3 + 1/√3.

Let me consider combining all terms into one fraction. However, since there are different radicals, it's not straightforward. Alternatively, maybe combining terms step by step.

Wait, but I think my initial approach is correct. Let me think again.

Alternatively, maybe I can compute the entire expression numerically, approximate the value, and check against (9√2 + 8√3)/6 to see if they match. That might help confirm.

Let me compute the original expression numerically.

Compute √2 ≈ 1.4142, 1/√2 ≈ 0.7071, √3 ≈ 1.732, 1/√3 ≈ 0.5774.

Adding them up: 1.4142 + 0.7071 + 1.732 + 0.5774 ≈ Let's compute:

First two terms: 1.4142 + 0.7071 ≈ 2.1213

Next two terms: 1.732 + 0.5774 ≈ 2.3094

Total ≈ 2.1213 + 2.3094 ≈ 4.4307

Now compute (9√2 + 8√3)/6:

First compute 9√2 ≈ 9 * 1.4142 ≈ 12.7278

Compute 8√3 ≈ 8 * 1.732 ≈ 13.856

Adding these: 12.7278 + 13.856 ≈ 26.5838

Divide by 6: 26.5838 / 6 ≈ 4.4306

Which is very close to the original sum of approximately 4.4307. So that's a good check. Therefore, my algebra seems correct.

Therefore, the answer should be 9 + 8 + 6 = 23. However, wait, the problem says "c is as small as possible". Let me check if there is a way to write this with a smaller denominator. Suppose someone tried to combine terms differently?

Wait, suppose that instead of rationalizing denominators first, maybe there's another approach? Let me think.

Alternatively, suppose we didn't rationalize denominators. Let me try that approach.

Original expression: √2 + 1/√2 + √3 + 1/√3.

Let me combine the √2 terms and the √3 terms. So for √2 terms: √2 + 1/√2. Let me write them with common denominator.

√2 can be written as (√2 * √2)/√2 = 2/√2. Wait, that's another way to think of it. Wait, √2 + 1/√2 = 2/√2 + 1/√2 = 3/√2. Wait, that's different from before. Wait, hold on. Wait, is that correct?

Wait, if I take √2 + 1/√2, if I write √2 as 2/√2, then adding 2/√2 + 1/√2 = 3/√2. But earlier, when I rationalized, I had 3√2/2. Wait, but 3/√2 is equal to 3√2/2. Because multiplying numerator and denominator by √2:

3/√2 = 3√2 / 2. So that's the same as before. Similarly, for the √3 terms: √3 + 1/√3 = 3/√3 = 3√3 / 3 = √3. Wait, hold on, that seems different. Wait, let me check again.

Wait, √3 + 1/√3. Let me again consider writing √3 as (√3 * √3)/√3 = 3/√3. Therefore, adding 3/√3 + 1/√3 = 4/√3. Then rationalizing gives 4√3 / 3. Which is the same as earlier. So that approach gives me √2 + 1/√2 = 3/√2 = 3√2 / 2 and √3 + 1/√3 = 4/√3 = 4√3 / 3. Then adding those together gives same result as before: 3√2 / 2 + 4√3 / 3. Therefore, same as before. So that approach also leads to same result. So, same as before.

Therefore, combining gives (9√2 + 8√3)/6, which seems correct. Therefore, I think my answer is correct. Therefore, a + b + c = 9 + 8 + 6 = 23.

Wait, but let me check if maybe there's a different way to combine terms where the denominator is smaller. For example, maybe there is a common factor that I missed? Let me think. Suppose that when I combined the two fractions (3√2)/2 + (4√3)/3, I converted them to have denominator 6. But maybe if there's another way to combine them?

Alternatively, suppose that if I didn't use denominators 2 and 3 but tried to combine them in another way? But I don't think so. The denominators are 2 and 3, so the LCD is 6. So that's the minimal denominator. So the fraction (9√2 + 8√3)/6 is already in simplest terms because 9, 8, and 6 have no common divisor except 1. Wait, 9 and 6 share 3 as a common factor, but since the numerator has 9√2 + 8√3, if we factor out 3 from the numerator, we would get 3*(3√2) + 8√3. But since the 8√3 term doesn't have a factor of 3, we can't factor 3 out of the entire numerator. Therefore, we can't reduce the fraction by dividing numerator and denominator by 3. Similarly, trying to divide by 2: 9 is odd, so can't divide by 2. Similarly with 8 and 6. Therefore, the fraction is reduced completely. Therefore, c = 6 is minimal.

Therefore, the answer should be 9 + 8 + 6 = 23. Wait, but let me check again with another approach. Let me check if there's another way to combine the original expression.

Alternatively, maybe I can combine all terms over a common denominator? Let me try that approach. Let me consider the entire expression:

√2 + 1/√2 + √3 + 1/√3. Let me think of combining all terms over a common denominator. But since there are different radicals, it's not straightforward. Let me try:

First, let me consider the entire expression as:

(√2 + 1/√2) + (√3 + 1/√3). Which is the same as before. Which we have already converted to 3√2/2 + 4√3/3. So, same result. Therefore, same as before.

Alternatively, suppose I tried to write all terms with denominator 6? Let me see.

Original terms:

√2 can be written as (6√2)/6.

But that seems unnecessary. Alternatively, perhaps not helpful. Alternatively, maybe think of combining the entire expression as follows:

Let me consider the entire expression as:

√2 + √3 + 1/√2 + 1/√3. Maybe combine the terms in pairs?

But that is again same as what I did before. So, seems same steps.

Alternatively, let me check if there is an error in my calculation of combining the terms. Let me recheck:

Original expression after rationalizing:

√2 + √2/2 + √3 + √3/3.

So for √2 terms: √2 + √2/2. √2 is 2√2/2. Then 2√2/2 + √2/2 = 3√2/2. Correct.

For √3 terms: √3 + √3/3. √3 is 3√3/3. Then 3√3/3 + √3/3 = 4√3/3. Correct. Then adding those two fractions: 3√2/2 + 4√3/3. To add them, common denominator is 6. Multiply first by 3/3: 9√2/6. Multiply second by 2/2: 8√3/6. So total is (9√2 + 8√3)/6. Correct.

Therefore, all steps check out. Therefore, the answer is 9 + 8 + 6 = 23.

But wait, just to be 100% sure, let me check if there's a way to have a smaller denominator. For example, suppose that there is some cancellation I haven't noticed? Let me think. Suppose that I thought of the entire original expression as:

Let me think of combining all terms over a common denominator. Let me think of all terms as fractions. Let me write √2 as √2/1, 1/√2 as 1/√2, √3 as √3/1, and 1/√3 as 1/√3.

But to combine them, I would need a common denominator. However, since they involve different radicals, the common denominator would be the product of denominators? Not sure. Alternatively, perhaps think of combining them step by step.

Alternatively, let me consider that maybe there's a mistake in assuming that we need to combine all terms. Wait, the problem says "in the form (a√2 + b√3)/c". So the numerator must be a combination of √2 and √3, and the denominator is c. Therefore, my previous approach is valid.

Alternatively, suppose that maybe there is another way to write the original expression with different a, b, c. For example, suppose that someone tried to combine √2 + 1/√2 + √3 + 1/√3 as follows:

Let me compute √2 + 1/√2. As I did before, that equals 3√2/2, and similarly √3 + 1/√3 is 4√3/3. Therefore, adding those gives 3√2/2 + 4√3/3. If I wanted to write this as a single fraction, it's (9√2 + 8√3)/6. So that's the only way? Or is there a different way?

Alternatively, if I write it as (3√2)/2 + (4√3)/3. Suppose I thought of writing them with denominator 6. Which is what I did. So that's the minimal denominator. Therefore, I think that is correct.

Alternatively, maybe there's a mistake in thinking that the numerator is 9√2 + 8√3. Let me check that again. For 3√2/2 converted to 6 denominator: multiply numerator and denominator by 3: 3√2 * 3 = 9√2, denominator 2*3=6. Similarly, 4√3/3 converted to 6 denominator: multiply numerator and denominator by 2: 4√3 * 2 = 8√3, denominator 3*2=6. Therefore, yes, the numerator is indeed 9√2 + 8√3 and denominator is 6. So that seems correct.

Alternatively, suppose someone thought of combining the original expression differently? For example, let me check:

Suppose I consider combining √2 + √3 first, and then 1/√2 + 1/√3. Let me see:

√2 + √3 + 1/√2 + 1/√3. Which is same as before. So same result. Alternatively, perhaps combining terms in another way? For instance, grouping √2 + 1/√2 and √3 + 1/√3 as done before. So same steps.

Alternatively, maybe trying to combine √2 + 1/√2 as (sqrt(2) * sqrt(2) + 1)/sqrt(2) = (2 + 1)/sqrt(2) = 3/sqrt(2). Which is same as before. So that gives 3/sqrt(2) + 4/sqrt(3). Then, combining these two terms over a common denominator? Wait, but they have different radicals, so how do you combine them? You can't combine different radicals unless you have like terms. Therefore, you need to rationalize them and then combine. So which is the same as before.

Therefore, I don't think there's another way. Therefore, the answer must be 23.

Wait, but let me check once more if there's a possible mistake in my calculation of the coefficients. For example, if I miscalculated when converting to denominator 6.

Original fractions:

First term: 3√2 / 2. To convert to denominator 6, multiply numerator and denominator by 3: numerator becomes 3√2 * 3 = 9√2. Denominator becomes 2 * 3 = 6. Correct.

Second term: 4√3 / 3. Multiply numerator and denominator by 2: numerator is 4√3 * 2 = 8√3. Denominator is 3 * 2 = 6. Correct.

Therefore, adding them gives 9√2 + 8√3 over 6. Correct. So that seems right.

Another check: Let me take a different approach. Let me compute the original expression and then compute (9√2 + 8√3)/6 and check if they are equal, as I did earlier. Since both give approximately 4.4306, which matches, that seems correct. Therefore, I think my answer is correct.

Therefore, the answer is 9 + 8 + 6 = 23.

But hold on, let me check once again if there is a way to write it with smaller c. For example, if I have (9√2 + 8√3)/6, can I divide numerator and denominator by something?

But as I thought earlier, the coefficients in numerator are 9 and 8. The GCD of 9 and 8 is 1, so there is no common factor between them. However, 6 is the denominator. If there was a common factor between denominator and numerator coefficients, we could reduce. However, since 6 and 9 share a common factor of 3, but since 8 and 3 are co-prime, we can't factor out 3. For example, if we try to divide numerator and denominator by 3, we would have:

(9√2 + 8√3)/6 = [3*(3√2) + 8√3]/6. Which is not a single term, so you cannot factor out 3 from the entire numerator. Therefore, you cannot reduce the fraction. Therefore, c = 6 is indeed minimal.

Alternatively, suppose someone tried to write the expression as something else. For example, suppose we thought of combining terms differently. For example, maybe adding √2 + √3 first. Let's see:

√2 + √3 + 1/√2 + 1/√3. Let me compute √2 + √3 ≈ 1.4142 + 1.732 ≈ 3.1462. And 1/√2 + 1/√3 ≈ 0.7071 + 0.5774 ≈ 1.2845. Then total ≈ 3.1462 + 1.2845 ≈ 4.4307, same as before. But this approach doesn't help us combine the terms into the desired form. So I think that my previous steps are correct.

Therefore, I think that my answer of 23 is correct. Therefore, I will go with that.

**Final Answer**
\boxed{23}

To simplify the expression:

$$
\sqrt{2} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}
$$

into the form:

$$
\frac{a\sqrt{2} + b\sqrt{3}}{c}
$$

where $ a, b, c $ are positive integers and $ c $ is as small as possible, we proceed step by step.

---

### **Step 1: Rationalize the Denominators**

We first rationalize the denominators of the fractional terms.

- $ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $
- $ \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $

Now the entire expression becomes:

$$
\sqrt{2} + \frac{\sqrt{2}}{2} + \sqrt{3} + \frac{\sqrt{3}}{3}
$$

---

### **Step 2: Combine Like Terms**

Group and combine the terms involving $ \sqrt{2} $ and $ \sqrt{3} $:

#### Combine $ \sqrt{2} $-terms:

$$
\sqrt{2} + \frac{\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}
$$

#### Combine $ \sqrt{3} $-terms:

$$
\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{3\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}
$$

---

### **Step 3: Combine the Results into a Single Fraction**

Now we add the two fractions:

$$
\frac{3\sqrt{2}}{2} + \frac{4\sqrt{3}}{3}
$$

Find the least common denominator of 2 and 3, which is 6. Convert both fractions:

$$
\frac{3\sqrt{2}}{2} = \frac{9\sqrt{2}}{6}, \quad \frac{4\sqrt{3}}{3} = \frac{8\sqrt{3}}{6}
$$

So the combined expression becomes:

$$
\frac{9\sqrt{2} + 8\sqrt{3}}{6}
$$

---

### **Step 4: Check for Simplification**

Now we check whether $ \frac{9\sqrt{2} + 8\sqrt{3}}{6} $ can be simplified further.

- The coefficients in the numerator are 9 and 8, which have no common factors.
- The denominator is 6.

So the greatest common divisor (GCD) of 9, 8, and 6 is 1. Therefore, the fraction is already in its simplest form.

Thus, we identify:

- $ a = 9 $
- $ b = 8 $
- $ c = 6 $

---

### **Final Step: Compute $ a + b + c $**

$$
a + b + c = 9 + 8 + 6 = \boxed{23}
$$

[Assistant]
Explanation:
Okay, let me try to figure out how to solve this problem. The question is asking me to write the expression √2 + 1/√2 + √3 + 1/√3 in the form (a√2 + b√3)/c, where a, b, and c are positive integers, and c is as small as possible. Then, I need to find a + b + c. Alright, let me start by breaking down the problem step by step.

First, let me write down the original expression:

√2 + 1/√2 + √3 + 1/√3.

Hmm, I need to combine these terms into a single fraction with the form (a√2 + b√3)/c. To do that, I probably need to simplify each term first, maybe rationalize the denominators where necessary. Let me recall how to rationalize denominators. For example, 1/√2 can be rationalized by multiplying numerator and denominator by √2, right?

Let me try that. Let me handle each term one by one.

Starting with √2. Well, that's already simplified. Then 1/√2. If I rationalize that, it becomes √2/2. Similarly, for 1/√3, rationalizing gives √3/3.

So let me rewrite the entire expression with these rationalized forms:

√2 + √2/2 + √3 + √3/3.

Okay, now I can combine like terms. Let me group the terms with √2 and the terms with √3.

For the √2 terms: √2 + √2/2. Let me compute that. To add these, I need a common denominator. The denominators here are 1 and 2, so common denominator is 2. So √2 is equal to 2√2/2. Therefore, adding 2√2/2 + √2/2 gives (2√2 + √2)/2 = 3√2/2.

Similarly, for the √3 terms: √3 + √3/3. Again, common denominator for 1 and 3 is 3. So √3 is 3√3/3. Therefore, adding 3√3/3 + √3/3 gives (3√3 + √3)/3 = 4√3/3.

So now the entire expression simplifies to 3√2/2 + 4√3/3. Now, the problem wants this in the form (a√2 + b√3)/c. So right now, I have two separate fractions: 3√2/2 and 4√3/3. To combine them into a single fraction, I need to find a common denominator. The denominators are 2 and 3, so the least common denominator is 6. Let me convert both fractions to have denominator 6.

First term: 3√2/2. Multiply numerator and denominator by 3: (3√2 * 3)/(2 * 3) = 9√2/6.

Second term: 4√3/3. Multiply numerator and denominator by 2: (4√3 * 2)/(3 * 2) = 8√3/6.

Now, adding these two fractions together: 9√2/6 + 8√3/6 = (9√2 + 8√3)/6.

So the expression simplifies to (9√2 + 8√3)/6. Now, check if this is in the desired form. The problem says that a, b, c should be positive integers and c is as small as possible. Here, a is 9, b is 8, and c is 6. Let me check if this fraction can be simplified further. For that, I need to see if 9, 8, and 6 have a common factor.

Looking at 9, 8, and 6. The factors:

- 9 factors: 1, 3, 9

- 8 factors: 1, 2, 4, 8

- 6 factors: 1, 2, 3, 6

The greatest common divisor (GCD) of all three numbers 9, 8, and 6. The GCD of 9 and 8 is 1, since they are co-prime. Therefore, the GCD of all three numbers is 1. Therefore, the fraction (9√2 + 8√3)/6 cannot be simplified further. Therefore, c is 6, which is already as small as possible.

Therefore, a = 9, b = 8, c = 6. Therefore, a + b + c = 9 + 8 + 6 = 23. Is that correct? Let me double-check my steps to make sure I didn't make any mistakes.

Starting again: Original expression √2 + 1/√2 + √3 + 1/√3. Rationalizing denominators:

√2 + √2/2 + √3 + √3/3. Combining √2 terms: √2 + √2/2 = (2√2 + √2)/2 = 3√2/2. Similarly for √3 terms: √3 + √3/3 = (3√3 + √3)/3 = 4√3/3. Then adding those two fractions: 3√2/2 + 4√3/3. Common denominator 6: 9√2/6 + 8√3/6 = (9√2 + 8√3)/6. So that seems correct. GCD of numerator coefficients (9,8,6) is 1, so c is indeed 6. So a + b + c is 9 + 8 + 6 = 23. Hmm.

Wait, but let me check if there's another way to combine the original terms that might lead to a different form with smaller c. Maybe I made a mistake in assuming that combining the terms as I did is the only way? Let me think.

Alternatively, perhaps I can combine all the terms over a common denominator from the start? Let me try that approach.

Original expression:

√2 + 1/√2 + √3 + 1/√3.

Let me consider combining all terms into one fraction. However, since there are different radicals, it's not straightforward. Alternatively, maybe combining terms step by step.

Wait, but I think my initial approach is correct. Let me think again.

Alternatively, maybe I can compute the entire expression numerically, approximate the value, and check against (9√2 + 8√3)/6 to see if they match. That might help confirm.

Let me compute the original expression numerically.

Compute √2 ≈ 1.4142, 1/√2 ≈ 0.7071, √3 ≈ 1.732, 1/√3 ≈ 0.5774.

Adding them up: 1.4142 + 0.7071 + 1.732 + 0.5774 ≈ Let's compute:

First two terms: 1.4142 + 0.7071 ≈ 2.1213

Next two terms: 1.732 + 0.5774 ≈ 2.3094

Total ≈ 2.1213 + 2.3094 ≈ 4.4307

Now compute (9√2 + 8√3)/6:

First compute 9√2 ≈ 9 * 1.4142 ≈ 12.7278

Compute 8√3 ≈ 8 * 1.732 ≈ 13.856

Adding these: 12.7278 + 13.856 ≈ 26.5838

Divide by 6: 26.5838 / 6 ≈ 4.4306

Which is very close to the original sum of approximately 4.4307. So that's a good check. Therefore, my algebra seems correct.

Therefore, the answer should be 9 + 8 + 6 = 23. However, wait, the problem says "c is as small as possible". Let me check if there is a way to write this with a smaller denominator. Suppose someone tried to combine terms differently?

Wait, suppose that instead of rationalizing denominators first, maybe there's another approach? Let me think.

Alternatively, suppose we didn't rationalize denominators. Let me try that approach.

Original expression: √2 + 1/√2 + √3 + 1/√3.

Let me combine the √2 terms and the √3 terms. So for √2 terms: √2 + 1/√2. Let me write them with common denominator.

√2 can be written as (√2 * √2)/√2 = 2/√2. Wait, that's another way to think of it. Wait, √2 + 1/√2 = 2/√2 + 1/√2 = 3/√2. Wait, that's different from before. Wait, hold on. Wait, is that correct?

Wait, if I take √2 + 1/√2, if I write √2 as 2/√2, then adding 2/√2 + 1/√2 = 3/√2. But earlier, when I rationalized, I had 3√2/2. Wait, but 3/√2 is equal to 3√2/2. Because multiplying numerator and denominator by √2:

3/√2 = 3√2 / 2. So that's the same as before. Similarly, for the √3 terms: √3 + 1/√3 = 3/√3 = 3√3 / 3 = √3. Wait, hold on, that seems different. Wait, let me check again.

Wait, √3 + 1/√3. Let me again consider writing √3 as (√3 * √3)/√3 = 3/√3. Therefore, adding 3/√3 + 1/√3 = 4/√3. Then rationalizing gives 4√3 / 3. Which is the same as earlier. So that approach gives me √2 + 1/√2 = 3/√2 = 3√2 / 2 and √3 + 1/√3 = 4/√3 = 4√3 / 3. Then adding those together gives same result as before: 3√2 / 2 + 4√3 / 3. Therefore, same as before. So that approach also leads to same result. So, same as before.

Therefore, combining gives (9√2 + 8√3)/6, which seems correct. Therefore, I think my answer is correct. Therefore, a + b + c = 9 + 8 + 6 = 23.

Wait, but let me check if maybe there's a different way to combine terms where the denominator is smaller. For example, maybe there is a common factor that I missed? Let me think. Suppose that when I combined the two fractions (3√2)/2 + (4√3)/3, I converted them to have denominator 6. But maybe if there's another way to combine them?

Alternatively, suppose that if I didn't use denominators 2 and 3 but tried to combine them in another way? But I don't think so. The denominators are 2 and 3, so the LCD is 6. So that's the minimal denominator. So the fraction (9√2 + 8√3)/6 is already in simplest terms because 9, 8, and 6 have no common divisor except 1. Wait, 9 and 6 share 3 as a common factor, but since the numerator has 9√2 + 8√3, if we factor out 3 from the numerator, we would get 3*(3√2) + 8√3. But since the 8√3 term doesn't have a factor of 3, we can't factor 3 out of the entire numerator. Therefore, we can't reduce the fraction by dividing numerator and denominator by 3. Similarly, trying to divide by 2: 9 is odd, so can't divide by 2. Similarly with 8 and 6. Therefore, the fraction is reduced completely. Therefore, c = 6 is minimal.

Therefore, the answer should be 9 + 8 + 6 = 23. Wait, but let me check again with another approach. Let me check if there's another way to combine the original expression.

Alternatively, maybe I can combine all terms over a common denominator? Let me try that approach. Let me consider the entire expression:

√2 + 1/√2 + √3 + 1/√3. Let me think of combining all terms over a common denominator. But since there are different radicals, it's not straightforward. Let me try:

First, let me consider the entire expression as:

(√2 + 1/√2) + (√3 + 1/√3). Which is the same as before. Which we have already converted to 3√2/2 + 4√3/3. So, same result. Therefore, same as before.

Alternatively, suppose I tried to write all terms with denominator 6? Let me see.

Original terms:

√2 can be written as (6√2)/6.

But that seems unnecessary. Alternatively, perhaps not helpful. Alternatively, maybe think of combining the entire expression as follows:

Let me consider the entire expression as:

√2 + √3 + 1/√2 + 1/√3. Maybe combine the terms in pairs?

But that is again same as what I did before. So, seems same steps.

Alternatively, let me check if there is an error in my calculation of combining the terms. Let me recheck:

Original expression after rationalizing:

√2 + √2/2 + √3 + √3/3.

So for √2 terms: √2 + √2/2. √2 is 2√2/2. Then 2√2/2 + √2/2 = 3√2/2. Correct.

For √3 terms: √3 + √3/3. √3 is 3√3/3. Then 3√3/3 + √3/3 = 4√3/3. Correct. Then adding those two fractions: 3√2/2 + 4√3/3. To add them, common denominator is 6. Multiply first by 3/3: 9√2/6. Multiply second by 2/2: 8√3/6. So total is (9√2 + 8√3)/6. Correct.

Therefore, all steps check out. Therefore, the answer is 9 + 8 + 6 = 23.

But wait, just to be 100% sure, let me check if there's a way to have a smaller denominator. For example, suppose that there is some cancellation I haven't noticed? Let me think. Suppose that I thought of the entire original expression as:

Let me think of combining all terms over a common denominator. Let me think of all terms as fractions. Let me write √2 as √2/1, 1/√2 as 1/√2, √3 as √3/1, and 1/√3 as 1/√3.

But to combine them, I would need a common denominator. However, since they involve different radicals, the common denominator would be the product of denominators? Not sure. Alternatively, perhaps think of combining them step by step.

Alternatively, let me consider that maybe there's a mistake in assuming that we need to combine all terms. Wait, the problem says "in the form (a√2 + b√3)/c". So the numerator must be a combination of √2 and √3, and the denominator is c. Therefore, my previous approach is valid.

Alternatively, suppose that maybe there is another way to write the original expression with different a, b, c. For example, suppose that someone tried to combine √2 + 1/√2 + √3 + 1/√3 as follows:

Let me compute √2 + 1/√2. As I did before, that equals 3√2/2, and similarly √3 + 1/√3 is 4√3/3. Therefore, adding those gives 3√2/2 + 4√3/3. If I wanted to write this as a single fraction, it's (9√2 + 8√3)/6. So that's the only way? Or is there a different way?

Alternatively, if I write it as (3√2)/2 + (4√3)/3. Suppose I thought of writing them with denominator 6. Which is what I did. So that's the minimal denominator. Therefore, I think that is correct.

Alternatively, maybe there's a mistake in thinking that the numerator is 9√2 + 8√3. Let me check that again. For 3√2/2 converted to 6 denominator: multiply numerator and denominator by 3: 3√2 * 3 = 9√2, denominator 2*3=6. Similarly, 4√3/3 converted to 6 denominator: multiply numerator and denominator by 2: 4√3 * 2 = 8√3, denominator 3*2=6. Therefore, yes, the numerator is indeed 9√2 + 8√3 and denominator is 6. So that seems correct.

Alternatively, suppose someone thought of combining the original expression differently? For example, let me check:

Suppose I consider combining √2 + √3 first, and then 1/√2 + 1/√3. Let me see:

√2 + √3 + 1/√2 + 1/√3. Which is same as before. So same result. Alternatively, perhaps combining terms in another way? For instance, grouping √2 + 1/√2 and √3 + 1/√3 as done before. So same steps.

Alternatively, maybe trying to combine √2 + 1/√2 as (sqrt(2) * sqrt(2) + 1)/sqrt(2) = (2 + 1)/sqrt(2) = 3/sqrt(2). Which is same as before. So that gives 3/sqrt(2) + 4/sqrt(3). Then, combining these two terms over a common denominator? Wait, but they have different radicals, so how do you combine them? You can't combine different radicals unless you have like terms. Therefore, you need to rationalize them and then combine. So which is the same as before.

Therefore, I don't think there's another way. Therefore, the answer must be 23.

Wait, but let me check once more if there's a possible mistake in my calculation of the coefficients. For example, if I miscalculated when converting to denominator 6.

Original fractions:

First term: 3√2 / 2. To convert to denominator 6, multiply numerator and denominator by 3: numerator becomes 3√2 * 3 = 9√2. Denominator becomes 2 * 3 = 6. Correct.

Second term: 4√3 / 3. Multiply numerator and denominator by 2: numerator is 4√3 * 2 = 8√3. Denominator is 3 * 2 = 6. Correct.

Therefore, adding them gives 9√2 + 8√3 over 6. Correct. So that seems right.

Another check: Let me take a different approach. Let me compute the original expression and then compute (9√2 + 8√3)/6 and check if they are equal, as I did earlier. Since both give approximately 4.4306, which matches, that seems correct. Therefore, I think my answer is correct.

Therefore, the answer is 9 + 8 + 6 = 23.

But hold on, let me check once again if there is a way to write it with smaller c. For example, if I have (9√2 + 8√3)/6, can I divide numerator and denominator by something?

But as I thought earlier, the coefficients in numerator are 9 and 8. The GCD of 9 and 8 is 1, so there is no common factor between them. However, 6 is the denominator. If there was a common factor between denominator and numerator coefficients, we could reduce. However, since 6 and 9 share a common factor of 3, but since 8 and 3 are co-prime, we can't factor out 3. For example, if we try to divide numerator and denominator by 3, we would have:

(9√2 + 8√3)/6 = [3*(3√2) + 8√3]/6. Which is not a single term, so you cannot factor out 3 from the entire numerator. Therefore, you cannot reduce the fraction. Therefore, c = 6 is indeed minimal.

Alternatively, suppose someone tried to write the expression as something else. For example, suppose we thought of combining terms differently. For example, maybe adding √2 + √3 first. Let's see:

√2 + √3 + 1/√2 + 1/√3. Let me compute √2 + √3 ≈ 1.4142 + 1.732 ≈ 3.1462. And 1/√2 + 1/√3 ≈ 0.7071 + 0.5774 ≈ 1.2845. Then total ≈ 3.1462 + 1.2845 ≈ 4.4307, same as before. But this approach doesn't help us combine the terms into the desired form. So I think that my previous steps are correct.

Therefore, I think that my answer of 23 is correct. Therefore, I will go with that.

**Final Answer**
\boxed{23}

To simplify the expression:

$$
\sqrt{2} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}
$$

into the form:

$$
\frac{a\sqrt{2} + b\sqrt{3}}{c}
$$

where $ a, b, c $ are positive integers and $ c $ is as small as possible, we proceed step by step.

---

### **Step 1: Rationalize the Denominators**

We first rationalize the denominators of the fractional terms.

- $ \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} $
- $ \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $

Now the entire expression becomes:

$$
\sqrt{2} + \frac{\sqrt{2}}{2} + \sqrt{3} + \frac{\sqrt{3}}{3}
$$

---

### **Step 2: Combine Like Terms**

Group and combine the terms involving $ \sqrt{2} $ and $ \sqrt{3} $:

#### Combine $ \sqrt{2} $-terms:

$$
\sqrt{2} + \frac{\sqrt{2}}{2} = \frac{2\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2}
$$

#### Combine $ \sqrt{3} $-terms:

$$
\sqrt{3} + \frac{\sqrt{3}}{3} = \frac{3\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}
$$

---

### **Step 3: Combine the Results into a Single Fraction**

Now we add the two fractions:

$$
\frac{3\sqrt{2}}{2} + \frac{4\sqrt{3}}{3}
$$

Find the least common denominator of 2 and 3, which is 6. Convert both fractions:

$$
\frac{3\sqrt{2}}{2} = \frac{9\sqrt{2}}{6}, \quad \frac{4\sqrt{3}}{3} = \frac{8\sqrt{3}}{6}
$$

So the combined expression becomes:

$$
\frac{9\sqrt{2} + 8\sqrt{3}}{6}
$$

---

### **Step 4: Check for Simplification**

Now we check whether $ \frac{9\sqrt{2} + 8\sqrt{3}}{6} $ can be simplified further.

- The coefficients in the numerator are 9 and 8, which have no common factors.
- The denominator is 6.

So the greatest common divisor (GCD) of 9, 8, and 6 is 1. Therefore, the fraction is already in its simplest form.

Thus, we identify:

- $ a = 9 $
- $ b = 8 $
- $ c = 6 $

---

### **Final Step: Compute $ a + b + c $**

$$
a + b + c = 9 + 8 + 6 = \boxed{23}
$$

Solution: The common denominator desired is $\sqrt{2}\cdot\sqrt{3} = \sqrt{6}$. So, this expression becomes $\frac{\sqrt{2}\cdot(\sqrt{2}\cdot\sqrt{3})+1\cdot\sqrt{3}+\sqrt{3}\cdot(\sqrt{2}\cdot\sqrt{3})+1\cdot\sqrt{2}}{\sqrt{6}}$. Simplifying this gives $\frac{2\sqrt{3}+\sqrt{3}+3\sqrt{2}+\sqrt{2}}{\sqrt{6}} = \frac{4\sqrt{2}+3\sqrt{3}}{\sqrt{6}}$. To rationalize, multiply numerator and denominator by $\sqrt{6}$ to get $\frac{4\sqrt{2}\sqrt{6}+3\sqrt{3}\sqrt{6}}{6}$. Simplifying yields ${\frac{9\sqrt{2}+8\sqrt{3}}{6}}$, so the desired sum is $9+8+6=\boxed{23}$.


Explanation:
Okay, so I need to figure out for how many positive integers n > 1 it's true that 2^24 is a perfect nth power. Hmm, let me start by recalling what a perfect nth power means. If a number is a perfect nth power, that means there exists some integer k such that k^n equals that number. In this case, the number is 2^24, so we need to find all n > 1 where there's an integer k with k^n = 2^24.

Let me think. Since 2^24 is already a power of 2, maybe I can express k as a power of 2 as well? Let me try that. Suppose k = 2^m for some integer m. Then k^n would be (2^m)^n = 2^{mn}. We want this to equal 2^24, so we need mn = 24. Therefore, n must be a divisor of 24, right? Because m and n are positive integers, and mn = 24, so n must divide 24.

But wait, the question says n > 1, so we need all divisors of 24 greater than 1. But hold on, is that all? Because maybe there are other possibilities where k isn't a power of 2? Let me check.

Suppose k isn't a power of 2. Then k would have some other prime factors. However, 2^24 only has 2 as a prime factor. So if k^n = 2^24, then k must also only have 2 as a prime factor. Otherwise, if k had another prime factor, say 3, then k^n would have 3^n as a factor, but 2^24 doesn't have any 3s. Therefore, k must be a power of 2. So my initial thought was correct. Therefore, n must divide 24, and n > 1.

Therefore, the problem reduces to finding the number of divisors of 24 that are greater than 1. Let me verify that.

First, let me factorize 24. 24 = 2^3 * 3^1. The number of positive divisors is (3 + 1)(1 + 1) = 4 * 2 = 8. Therefore, the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Excluding 1, we have 7 divisors. So the answer is 7? Wait, but hold on, let me make sure.

But wait, is there a case where n divides 24, but mn = 24, but m is not an integer? Wait, no. Since n divides 24, then m = 24/n, which is an integer. Because if n divides 24, then 24 is a multiple of n, so m is an integer. Therefore, for every divisor n of 24 greater than 1, there is such a k. Hence, the number of n is equal to the number of divisors of 24 greater than 1, which is 7. Therefore, the answer is 7?

Wait, but let me check with examples. Let's take n=2. Then k^2 = 2^24, so k=2^12. That works. Similarly, n=3: k^3=2^24, so k=2^8. Because 8*3=24. Similarly, n=4: k^4=2^24, so k=2^6. n=6: k^6=2^24, so k=2^4. n=8: k^8=2^24, so k=2^3. n=12: k^12=2^24, so k=2^2. n=24: k^24=2^24, so k=2^1. So all these work. So that's 7 values: 2,3,4,6,8,12,24. So 7 numbers. So the answer is 7? But wait, hold on. Let me check if there are other n's not divisors of 24 that can also satisfy the condition?

Suppose n is not a divisor of 24, but 24 is a multiple of n. Wait, no, if n divides 24, then 24 is a multiple of n. So if n doesn't divide 24, then 24 isn't a multiple of n, so mn=24 would not have integer m. Therefore, n must divide 24.

Alternatively, suppose someone thinks that maybe n could be a fraction? But n has to be a positive integer greater than 1. So fractions are out. So the conclusion is that n must be a divisor of 24 greater than 1, so 7 values. Therefore, answer is 7?

But wait, hold on. Let me check the divisors again. Divisors of 24 are:

1, 2, 3, 4, 6, 8, 12, 24. So removing 1, there are 7 divisors. Therefore, 7 values of n. Therefore, the answer is 7. But let me check again if there are other possibilities where k isn't just a power of 2. But as discussed earlier, since 2^24 only has prime factor 2, then k must be power of 2. Therefore, no other possibilities. Therefore, answer is 7. Is this correct?

Wait, but let me check if for example n= something else. Let me think. Suppose n= something like 5. Is 2^24 a perfect 5th power? Let me see. For 2^24 to be a perfect 5th power, the exponent 24 must be divisible by 5. But 24 divided by 5 is 4.8, which is not integer. Therefore, no. Similarly, n=7: 24 divided by 7 is not integer. So no. Therefore, only n that divide 24. Therefore, the answer is 7. Therefore, the number of positive integers n>1 is 7. So the answer is \boxed{7}?

But let me check once again. Maybe I made a mistake in the number of divisors. Let me list them again:

Divisors of 24:

Start with 1.

Then primes: 2, 3.

Then products:

2*2=4, 2*3=6, 2*2*2=8, 2*2*3=12, 2*2*2*3=24.

So divisors are 1, 2, 3, 4, 6, 8, 12, 24. That's 8 divisors. Subtract 1, gives 7. So that's correct.

Alternatively, maybe the question is trickier? Let me check the exact wording.

"For how many positive integers n>1 is it true that 2^24 is a perfect n^th power?"

Yes, so perfect nth power, meaning that there exists an integer k such that k^n = 2^24. Which as we said requires that n divides 24, so there are 7 such n. Therefore, the answer is 7. So I think that's correct. Maybe I need to check if there's another way to think about it?

Alternatively, think of exponents. 2^24 is equal to (2^a)^n. Therefore, 2^24 = 2^{a n}, so 24 = a n. Therefore, n must divide 24. So again, same conclusion. So n is a positive integer divisor of 24 greater than 1, so 7 numbers. Therefore, I think that is correct.

But just to be thorough, let me think of some other angle. Suppose someone says that maybe exponents can be fractions? But no, n is a positive integer, so exponents must be integers. So even if you think of roots, the nth root of 2^24 is 2^{24/n}, which needs to be integer, so 24/n must be integer. Therefore, n divides 24. So same result. Therefore, yes, the answer is 7.

So I think the answer is 7. Therefore, \boxed{7} is the answer.

**Final Answer**
\boxed{7}

To determine how many positive integers $ n > 1 $ satisfy the condition that $ 2^{24} $ is a perfect $ n $-th power, we must first understand what it means for a number to be a perfect $ n $-th power.

---

### Understanding the Problem

A number $ x $ is a **perfect $ n $-th power** if there exists some integer $ k $ such that:

$$
x = k^n
$$

In our case, $ x = 2^{24} $. So we are looking for all integers $ n > 1 $ such that there exists an integer $ k $ with:

$$
2^{24} = k^n
$$

---

### Analyzing the Structure

Since $ 2^{24} $ is a power of 2, let's suppose $ k = 2^m $ for some integer $ m $. Then:

$$
k^n = (2^m)^n = 2^{mn}
$$

We want this to equal $ 2^{24} $, so we must have:

$$
mn = 24
$$

Thus, every possible $ n > 1 $ such that $ n $ divides 24 yields a valid $ m = \frac{24}{n} $. Since $ m $ must be an integer, $ n $ must be a **positive divisor of 24**.

---

### Enumerating the Divisors of 24

The prime factorization of 24 is:

$$
24 = 2^3 \times 3^1
$$

Using the formula for the number of positive divisors:

$$
(3 + 1)(1 + 1) = 4 \times 2 = 8
$$

So 24 has 8 positive divisors:

$$
1, 2, 3, 4, 6, 8, 12, 24
$$

We are interested only in $ n > 1 $, so we remove 1:

$$
\text{Valid } n = 2, 3, 4, 6, 8, 12, 24
$$

This gives us **7** positive integers $ n > 1 $ such that $ 2^{24} $ is a perfect $ n $-th power.

---

### Final Answer

$$
\boxed{7}
$$

[Assistant]
Explanation:
Okay, so I need to figure out for how many positive integers n > 1 it's true that 2^24 is a perfect nth power. Hmm, let me start by recalling what a perfect nth power means. If a number is a perfect nth power, that means there exists some integer k such that k^n equals that number. In this case, the number is 2^24, so we need to find all n > 1 where there's an integer k with k^n = 2^24.

Let me think. Since 2^24 is already a power of 2, maybe I can express k as a power of 2 as well? Let me try that. Suppose k = 2^m for some integer m. Then k^n would be (2^m)^n = 2^{mn}. We want this to equal 2^24, so we need mn = 24. Therefore, n must be a divisor of 24, right? Because m and n are positive integers, and mn = 24, so n must divide 24.

But wait, the question says n > 1, so we need all divisors of 24 greater than 1. But hold on, is that all? Because maybe there are other possibilities where k isn't a power of 2? Let me check.

Suppose k isn't a power of 2. Then k would have some other prime factors. However, 2^24 only has 2 as a prime factor. So if k^n = 2^24, then k must also only have 2 as a prime factor. Otherwise, if k had another prime factor, say 3, then k^n would have 3^n as a factor, but 2^24 doesn't have any 3s. Therefore, k must be a power of 2. So my initial thought was correct. Therefore, n must divide 24, and n > 1.

Therefore, the problem reduces to finding the number of divisors of 24 that are greater than 1. Let me verify that.

First, let me factorize 24. 24 = 2^3 * 3^1. The number of positive divisors is (3 + 1)(1 + 1) = 4 * 2 = 8. Therefore, the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. Excluding 1, we have 7 divisors. So the answer is 7? Wait, but hold on, let me make sure.

But wait, is there a case where n divides 24, but mn = 24, but m is not an integer? Wait, no. Since n divides 24, then m = 24/n, which is an integer. Because if n divides 24, then 24 is a multiple of n, so m is an integer. Therefore, for every divisor n of 24 greater than 1, there is such a k. Hence, the number of n is equal to the number of divisors of 24 greater than 1, which is 7. Therefore, the answer is 7?

Wait, but let me check with examples. Let's take n=2. Then k^2 = 2^24, so k=2^12. That works. Similarly, n=3: k^3=2^24, so k=2^8. Because 8*3=24. Similarly, n=4: k^4=2^24, so k=2^6. n=6: k^6=2^24, so k=2^4. n=8: k^8=2^24, so k=2^3. n=12: k^12=2^24, so k=2^2. n=24: k^24=2^24, so k=2^1. So all these work. So that's 7 values: 2,3,4,6,8,12,24. So 7 numbers. So the answer is 7? But wait, hold on. Let me check if there are other n's not divisors of 24 that can also satisfy the condition?

Suppose n is not a divisor of 24, but 24 is a multiple of n. Wait, no, if n divides 24, then 24 is a multiple of n. So if n doesn't divide 24, then 24 isn't a multiple of n, so mn=24 would not have integer m. Therefore, n must divide 24.

Alternatively, suppose someone thinks that maybe n could be a fraction? But n has to be a positive integer greater than 1. So fractions are out. So the conclusion is that n must be a divisor of 24 greater than 1, so 7 values. Therefore, answer is 7?

But wait, hold on. Let me check the divisors again. Divisors of 24 are:

1, 2, 3, 4, 6, 8, 12, 24. So removing 1, there are 7 divisors. Therefore, 7 values of n. Therefore, the answer is 7. But let me check again if there are other possibilities where k isn't just a power of 2. But as discussed earlier, since 2^24 only has prime factor 2, then k must be power of 2. Therefore, no other possibilities. Therefore, answer is 7. Is this correct?

Wait, but let me check if for example n= something else. Let me think. Suppose n= something like 5. Is 2^24 a perfect 5th power? Let me see. For 2^24 to be a perfect 5th power, the exponent 24 must be divisible by 5. But 24 divided by 5 is 4.8, which is not integer. Therefore, no. Similarly, n=7: 24 divided by 7 is not integer. So no. Therefore, only n that divide 24. Therefore, the answer is 7. Therefore, the number of positive integers n>1 is 7. So the answer is \boxed{7}?

But let me check once again. Maybe I made a mistake in the number of divisors. Let me list them again:

Divisors of 24:

Start with 1.

Then primes: 2, 3.

Then products:

2*2=4, 2*3=6, 2*2*2=8, 2*2*3=12, 2*2*2*3=24.

So divisors are 1, 2, 3, 4, 6, 8, 12, 24. That's 8 divisors. Subtract 1, gives 7. So that's correct.

Alternatively, maybe the question is trickier? Let me check the exact wording.

"For how many positive integers n>1 is it true that 2^24 is a perfect n^th power?"

Yes, so perfect nth power, meaning that there exists an integer k such that k^n = 2^24. Which as we said requires that n divides 24, so there are 7 such n. Therefore, the answer is 7. So I think that's correct. Maybe I need to check if there's another way to think about it?

Alternatively, think of exponents. 2^24 is equal to (2^a)^n. Therefore, 2^24 = 2^{a n}, so 24 = a n. Therefore, n must divide 24. So again, same conclusion. So n is a positive integer divisor of 24 greater than 1, so 7 numbers. Therefore, I think that is correct.

But just to be thorough, let me think of some other angle. Suppose someone says that maybe exponents can be fractions? But no, n is a positive integer, so exponents must be integers. So even if you think of roots, the nth root of 2^24 is 2^{24/n}, which needs to be integer, so 24/n must be integer. Therefore, n divides 24. So same result. Therefore, yes, the answer is 7.

So I think the answer is 7. Therefore, \boxed{7} is the answer.

**Final Answer**
\boxed{7}

To determine how many positive integers $ n > 1 $ satisfy the condition that $ 2^{24} $ is a perfect $ n $-th power, we must first understand what it means for a number to be a perfect $ n $-th power.

---

### Understanding the Problem

A number $ x $ is a **perfect $ n $-th power** if there exists some integer $ k $ such that:

$$
x = k^n
$$

In our case, $ x = 2^{24} $. So we are looking for all integers $ n > 1 $ such that there exists an integer $ k $ with:

$$
2^{24} = k^n
$$

---

### Analyzing the Structure

Since $ 2^{24} $ is a power of 2, let's suppose $ k = 2^m $ for some integer $ m $. Then:

$$
k^n = (2^m)^n = 2^{mn}
$$

We want this to equal $ 2^{24} $, so we must have:

$$
mn = 24
$$

Thus, every possible $ n > 1 $ such that $ n $ divides 24 yields a valid $ m = \frac{24}{n} $. Since $ m $ must be an integer, $ n $ must be a **positive divisor of 24**.

---

### Enumerating the Divisors of 24

The prime factorization of 24 is:

$$
24 = 2^3 \times 3^1
$$

Using the formula for the number of positive divisors:

$$
(3 + 1)(1 + 1) = 4 \times 2 = 8
$$

So 24 has 8 positive divisors:

$$
1, 2, 3, 4, 6, 8, 12, 24
$$

We are interested only in $ n > 1 $, so we remove 1:

$$
\text{Valid } n = 2, 3, 4, 6, 8, 12, 24
$$

This gives us **7** positive integers $ n > 1 $ such that $ 2^{24} $ is a perfect $ n $-th power.

---

### Final Answer

$$
\boxed{7}
$$

Solution: Note that $2^{24}$ is a perfect $n$th power if and only if $n$ is a divisor of 24. The factors of 24 which are greater than 1 are 2, 3, 4, 6, 8, 12, and 24, so we have $\boxed{7}$ possible values of $n$.


Explanation:
Okay, let me try to figure out this problem. So, we have a monic polynomial p(x) of degree 4. Three of its roots are 1, 2, and 3. We need to find p(0) + p(4). Hmm, okay. Let's 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 fourth-degree polynomial, it has four roots (counting multiplicities). We know three of them: 1, 2, and 3. So there must be a fourth root, let's call it r.

So, the polynomial can be written in factored form as p(x) = (x - 1)(x - 2)(x - 3)(x - r). Since it's monic, the leading coefficient is 1, so that's correct. Now, we need to figure out what r is. Wait, but the problem doesn't give us any more information directly. Maybe there's something else we can use?

Wait, maybe we don't need to find r explicitly? Let me think. The problem asks for p(0) + p(4). Let me try to compute p(0) and p(4) using the factored form.

Starting with p(0). If we plug in x = 0 into the polynomial, we get:

p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - r) = (-1)(-2)(-3)(-r). Let me compute that step by step.

First, (-1) * (-2) = 2. Then, 2 * (-3) = -6. Then, -6 * (-r) = 6r. So p(0) = 6r.

Now, let's compute p(4). Plugging x = 4 into the polynomial:

p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - r) = (3)(2)(1)(4 - r). Let me compute that.

3 * 2 = 6, 6 * 1 = 6, so 6*(4 - r). Therefore, p(4) = 6*(4 - r) = 24 - 6r.

Therefore, p(0) + p(4) = 6r + 24 - 6r = 24. Wait, that's interesting! The 6r and -6r cancel out, so it's just 24? But that seems too straightforward. Did I do something wrong here? Let me check again.

Starting with p(0): (0 - 1)(0 - 2)(0 - 3)(0 - r). Let me compute each term:

(0 - 1) = -1

(0 - 2) = -2

(0 - 3) = -3

(0 - r) = -r

Multiply them all together: (-1) * (-2) * (-3) * (-r) = [(-1)*(-2)] * [(-3)*(-r)] = (2)*(3r) = 6r. That's correct.

For p(4):

(4 - 1) = 3

(4 - 2) = 2

(4 - 3) = 1

(4 - r) = 4 - r

Multiply them: 3*2*1*(4 - r) = 6*(4 - r) = 24 - 6r. Correct.

Then adding them together: 6r + 24 - 6r = 24. So, regardless of the value of r, p(0) + p(4) is always 24? That seems surprising. But since the fourth root cancels out, maybe that's the case?

Wait, but the problem didn't specify anything else about the polynomial. So, does that mean that regardless of the fourth root, p(0) + p(4) is always 24? That seems to be the case here. Is there a different approach to verify this?

Alternatively, maybe we can think of the polynomial as p(x) = (x - 1)(x - 2)(x - 3)(x - r), and expand it, then compute p(0) and p(4).

But expanding might be tedious, but maybe we can do it. Let me try. Let's first compute (x - 1)(x - 2)(x - 3). Let me compute that first.

First multiply (x - 1)(x - 2) = x² - 3x + 2. Then multiply by (x - 3):

(x² - 3x + 2)(x - 3) = x³ - 3x² - 3x² + 9x + 2x - 6. Wait, let me compute term by term:

Multiply x² by (x - 3): x³ - 3x².

Multiply -3x by (x - 3): -3x² + 9x.

Multiply 2 by (x - 3): 2x - 6.

Now, add them all together:

x³ - 3x² - 3x² + 9x + 2x - 6 = x³ - 6x² + 11x - 6.

Therefore, the polynomial p(x) is (x³ - 6x² + 11x - 6)(x - r). Let's expand that.

Multiply x³ by (x - r): x⁴ - r x³.

Multiply -6x² by (x - r): -6x³ + 6r x².

Multiply 11x by (x - r): 11x² - 11r x.

Multiply -6 by (x - r): -6x + 6r.

Now, add all these terms together:

x⁴ - r x³ - 6x³ + 6r x² + 11x² - 11r x - 6x + 6r.

Combine like terms:

x⁴ + (-r - 6)x³ + (6r + 11)x² + (-11r - 6)x + 6r.

So p(x) = x⁴ - (r + 6)x³ + (6r + 11)x² - (11r + 6)x + 6r.

Now, let's compute p(0). When x = 0, all terms with x will vanish, so p(0) is the constant term, which is 6r. Which matches what we had before.

Similarly, p(4) can be calculated by plugging in x = 4 into the expanded polynomial. Alternatively, since we already have p(4) as 24 - 6r, but let me check using the expanded form.

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

Let me compute each term step by step:

First term: 4^4 = 256.

Second term: -(r + 6)*4^3. 4^3 is 64. So, -(r + 6)*64 = -64r - 384.

Third term: (6r + 11)*4^2. 4^2 is 16. So, (6r + 11)*16 = 96r + 176.

Fourth term: -(11r + 6)*4 = -44r - 24.

Fifth term: 6r.

Now, combine all these terms:

256 -64r - 384 + 96r + 176 -44r -24 + 6r.

Let me combine like terms:

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

Compute step by step:

256 - 384 = -128.

-128 + 176 = 48.

48 - 24 = 24.

Now, the terms with r:

-64r + 96r -44r + 6r.

Compute step by step:

-64r + 96r = 32r.

32r -44r = -12r.

-12r + 6r = -6r.

So overall, p(4) is 24 - 6r, which matches what we had before. Therefore, p(0) + p(4) is 6r + 24 - 6r = 24.

So regardless of the value of r, which is the fourth root, the sum p(0) + p(4) is always 24. Therefore, the answer should be 24.

Wait, but that seems counterintuitive. How can the value not depend on the fourth root? Maybe there's some symmetry here? Let me think again.

Alternatively, maybe there's another way to see this. Let me think about Vieta's formula or something else.

But since the polynomial is monic, degree 4, with roots 1, 2, 3, and r. Then, if we consider p(0) is the product of the roots multiplied by (-1)^4, since for a monic polynomial, the constant term is (-1)^degree times the product of the roots. Since degree is 4, it's (-1)^4 * product of roots = product of roots. Therefore, p(0) = 1*2*3*r = 6r. Which matches our previous result.

Similarly, p(4) is the value of the polynomial at x = 4. But how does that relate? Well, maybe there's another way to think about p(4) in terms of the roots. However, it's probably more straightforward as we did before.

But the key point is that when we add p(0) + p(4), the term involving r cancels out. Therefore, regardless of r, the sum is fixed. Therefore, the answer is 24.

Is there any way that the problem is missing something? For example, maybe the polynomial is supposed to have distinct roots? But even if there is a repeated root, like r is one of 1, 2, 3, then the same formula would apply. For example, if r is 1, then p(x) would be (x - 1)^2(x - 2)(x - 3). Then p(0) would be ( -1)^2*(-2)*(-3) = 1*6 = 6. Wait, but according to our formula, p(0) = 6r, which would be 6*1 = 6. Correct. Then p(4) would be 6*(4 - 1) = 6*3 = 18. Then p(0) + p(4) would be 6 + 18 = 24. So same result.

Similarly, if r = 2, p(0) = 6*2 = 12. Then p(4) = 6*(4 - 2) = 6*2 = 12. Then total is 12 + 12 = 24. Same. If r = 3, p(0) = 6*3 = 18, p(4) = 6*(4 - 3) = 6*1 = 6, total is 18 + 6 = 24. If r = 0, p(0) = 6*0 = 0, p(4) = 6*(4 - 0) = 24, so 0 + 24 = 24.

So regardless of which root we take, even if it's a duplicate, the sum p(0) + p(4) is always 24. Therefore, the answer is 24.

But let me check with a concrete example. Let's choose r = 4. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 4). Let's compute p(0) + p(4).

Compute p(0): (-1)(-2)(-3)(-4) = (-1*-2) = 2, 2*-3 = -6, -6*-4 = 24. So p(0) = 24.

Compute p(4): (4 - 1)(4 - 2)(4 - 3)(4 - 4) = 3*2*1*0 = 0. Therefore, p(0) + p(4) = 24 + 0 = 24. Yep, same result.

Another example: let me take r = 5. Then p(0) = 6*5 = 30. p(4) = 6*(4 - 5) = 6*(-1) = -6. Then p(0) + p(4) = 30 - 6 = 24. Yep.

Therefore, regardless of the value of r, the sum is 24. Therefore, the answer is 24.

But wait, why does this happen? Is there a deeper reason? Maybe considering the polynomial's symmetry? Let me think.

Suppose we consider the polynomial p(x) with roots 1, 2, 3, r. Then, if we consider the function f(x) = p(x) * something... Maybe not. Alternatively, consider that p(0) is product of roots, which is 6r. And p(4) is... Well, when we plug in x=4, it's related to the value of the polynomial at 4. However, when we add p(0) and p(4), the terms with r cancel.

Alternatively, think of the polynomial as being symmetric in some way? For example, if we have roots symmetric around some point. But with roots at 1, 2, 3, and r, unless r is chosen such that there's symmetry. However, the fact that the sum p(0) + p(4) is constant regardless of r seems to be a coincidence of the specific form of the polynomial.

Alternatively, think of p(0) + p(4) as something involving the roots. But since p(0) = 6r and p(4) = 6(4 - r), adding them cancels the r term.

So maybe this is a general result? For example, if you have a monic polynomial with roots a, b, c, d, then p(0) is the product of the roots, and p(k) is something... But in this case, with the specific roots 1, 2, 3, and r, and evaluating at 0 and 4, the expression p(0) + p(4) ends up being constant.

But regardless, the math checks out with multiple examples and through two different methods (factoring and expanding). So I think the answer is indeed 24.

Therefore, I think the answer is \boxed{24}.

**Final Answer**
\boxed{24}

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

---

### Step 1: General Form of $ p(x) $

Since $ p(x) $ is a fourth-degree monic polynomial, and we know three of its roots, the general form must be:

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

where $ r $ is the unknown fourth root. This form guarantees that $ p(x) $ is monic, and all four roots are taken into account, counted with possible multiplicities.

---

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

We substitute $ x = 0 $ into the polynomial:

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

Multiply the factors:

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

So,

$$
p(0) = 6r
$$

---

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

Now substitute $ x = 4 $ into the same expression:

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

Multiply the constants:

$$
3\cdot2\cdot1 = 6,\quad 6\cdot(4 - r) = 24 - 6r
$$

So,

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

---

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

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

Simplify:

$$
6r - 6r + 24 = 24
$$

---

### Final Result

The value of $ p(0) + p(4) $ is **independent of the value of $ r $**, and always equals:

$$
\boxed{24}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, we have a monic polynomial p(x) of degree 4. Three of its roots are 1, 2, and 3. We need to find p(0) + p(4). Hmm, okay. Let's 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 fourth-degree polynomial, it has four roots (counting multiplicities). We know three of them: 1, 2, and 3. So there must be a fourth root, let's call it r.

So, the polynomial can be written in factored form as p(x) = (x - 1)(x - 2)(x - 3)(x - r). Since it's monic, the leading coefficient is 1, so that's correct. Now, we need to figure out what r is. Wait, but the problem doesn't give us any more information directly. Maybe there's something else we can use?

Wait, maybe we don't need to find r explicitly? Let me think. The problem asks for p(0) + p(4). Let me try to compute p(0) and p(4) using the factored form.

Starting with p(0). If we plug in x = 0 into the polynomial, we get:

p(0) = (0 - 1)(0 - 2)(0 - 3)(0 - r) = (-1)(-2)(-3)(-r). Let me compute that step by step.

First, (-1) * (-2) = 2. Then, 2 * (-3) = -6. Then, -6 * (-r) = 6r. So p(0) = 6r.

Now, let's compute p(4). Plugging x = 4 into the polynomial:

p(4) = (4 - 1)(4 - 2)(4 - 3)(4 - r) = (3)(2)(1)(4 - r). Let me compute that.

3 * 2 = 6, 6 * 1 = 6, so 6*(4 - r). Therefore, p(4) = 6*(4 - r) = 24 - 6r.

Therefore, p(0) + p(4) = 6r + 24 - 6r = 24. Wait, that's interesting! The 6r and -6r cancel out, so it's just 24? But that seems too straightforward. Did I do something wrong here? Let me check again.

Starting with p(0): (0 - 1)(0 - 2)(0 - 3)(0 - r). Let me compute each term:

(0 - 1) = -1

(0 - 2) = -2

(0 - 3) = -3

(0 - r) = -r

Multiply them all together: (-1) * (-2) * (-3) * (-r) = [(-1)*(-2)] * [(-3)*(-r)] = (2)*(3r) = 6r. That's correct.

For p(4):

(4 - 1) = 3

(4 - 2) = 2

(4 - 3) = 1

(4 - r) = 4 - r

Multiply them: 3*2*1*(4 - r) = 6*(4 - r) = 24 - 6r. Correct.

Then adding them together: 6r + 24 - 6r = 24. So, regardless of the value of r, p(0) + p(4) is always 24? That seems surprising. But since the fourth root cancels out, maybe that's the case?

Wait, but the problem didn't specify anything else about the polynomial. So, does that mean that regardless of the fourth root, p(0) + p(4) is always 24? That seems to be the case here. Is there a different approach to verify this?

Alternatively, maybe we can think of the polynomial as p(x) = (x - 1)(x - 2)(x - 3)(x - r), and expand it, then compute p(0) and p(4).

But expanding might be tedious, but maybe we can do it. Let me try. Let's first compute (x - 1)(x - 2)(x - 3). Let me compute that first.

First multiply (x - 1)(x - 2) = x² - 3x + 2. Then multiply by (x - 3):

(x² - 3x + 2)(x - 3) = x³ - 3x² - 3x² + 9x + 2x - 6. Wait, let me compute term by term:

Multiply x² by (x - 3): x³ - 3x².

Multiply -3x by (x - 3): -3x² + 9x.

Multiply 2 by (x - 3): 2x - 6.

Now, add them all together:

x³ - 3x² - 3x² + 9x + 2x - 6 = x³ - 6x² + 11x - 6.

Therefore, the polynomial p(x) is (x³ - 6x² + 11x - 6)(x - r). Let's expand that.

Multiply x³ by (x - r): x⁴ - r x³.

Multiply -6x² by (x - r): -6x³ + 6r x².

Multiply 11x by (x - r): 11x² - 11r x.

Multiply -6 by (x - r): -6x + 6r.

Now, add all these terms together:

x⁴ - r x³ - 6x³ + 6r x² + 11x² - 11r x - 6x + 6r.

Combine like terms:

x⁴ + (-r - 6)x³ + (6r + 11)x² + (-11r - 6)x + 6r.

So p(x) = x⁴ - (r + 6)x³ + (6r + 11)x² - (11r + 6)x + 6r.

Now, let's compute p(0). When x = 0, all terms with x will vanish, so p(0) is the constant term, which is 6r. Which matches what we had before.

Similarly, p(4) can be calculated by plugging in x = 4 into the expanded polynomial. Alternatively, since we already have p(4) as 24 - 6r, but let me check using the expanded form.

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

Let me compute each term step by step:

First term: 4^4 = 256.

Second term: -(r + 6)*4^3. 4^3 is 64. So, -(r + 6)*64 = -64r - 384.

Third term: (6r + 11)*4^2. 4^2 is 16. So, (6r + 11)*16 = 96r + 176.

Fourth term: -(11r + 6)*4 = -44r - 24.

Fifth term: 6r.

Now, combine all these terms:

256 -64r - 384 + 96r + 176 -44r -24 + 6r.

Let me combine like terms:

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

Compute step by step:

256 - 384 = -128.

-128 + 176 = 48.

48 - 24 = 24.

Now, the terms with r:

-64r + 96r -44r + 6r.

Compute step by step:

-64r + 96r = 32r.

32r -44r = -12r.

-12r + 6r = -6r.

So overall, p(4) is 24 - 6r, which matches what we had before. Therefore, p(0) + p(4) is 6r + 24 - 6r = 24.

So regardless of the value of r, which is the fourth root, the sum p(0) + p(4) is always 24. Therefore, the answer should be 24.

Wait, but that seems counterintuitive. How can the value not depend on the fourth root? Maybe there's some symmetry here? Let me think again.

Alternatively, maybe there's another way to see this. Let me think about Vieta's formula or something else.

But since the polynomial is monic, degree 4, with roots 1, 2, 3, and r. Then, if we consider p(0) is the product of the roots multiplied by (-1)^4, since for a monic polynomial, the constant term is (-1)^degree times the product of the roots. Since degree is 4, it's (-1)^4 * product of roots = product of roots. Therefore, p(0) = 1*2*3*r = 6r. Which matches our previous result.

Similarly, p(4) is the value of the polynomial at x = 4. But how does that relate? Well, maybe there's another way to think about p(4) in terms of the roots. However, it's probably more straightforward as we did before.

But the key point is that when we add p(0) + p(4), the term involving r cancels out. Therefore, regardless of r, the sum is fixed. Therefore, the answer is 24.

Is there any way that the problem is missing something? For example, maybe the polynomial is supposed to have distinct roots? But even if there is a repeated root, like r is one of 1, 2, 3, then the same formula would apply. For example, if r is 1, then p(x) would be (x - 1)^2(x - 2)(x - 3). Then p(0) would be ( -1)^2*(-2)*(-3) = 1*6 = 6. Wait, but according to our formula, p(0) = 6r, which would be 6*1 = 6. Correct. Then p(4) would be 6*(4 - 1) = 6*3 = 18. Then p(0) + p(4) would be 6 + 18 = 24. So same result.

Similarly, if r = 2, p(0) = 6*2 = 12. Then p(4) = 6*(4 - 2) = 6*2 = 12. Then total is 12 + 12 = 24. Same. If r = 3, p(0) = 6*3 = 18, p(4) = 6*(4 - 3) = 6*1 = 6, total is 18 + 6 = 24. If r = 0, p(0) = 6*0 = 0, p(4) = 6*(4 - 0) = 24, so 0 + 24 = 24.

So regardless of which root we take, even if it's a duplicate, the sum p(0) + p(4) is always 24. Therefore, the answer is 24.

But let me check with a concrete example. Let's choose r = 4. Then the polynomial is (x - 1)(x - 2)(x - 3)(x - 4). Let's compute p(0) + p(4).

Compute p(0): (-1)(-2)(-3)(-4) = (-1*-2) = 2, 2*-3 = -6, -6*-4 = 24. So p(0) = 24.

Compute p(4): (4 - 1)(4 - 2)(4 - 3)(4 - 4) = 3*2*1*0 = 0. Therefore, p(0) + p(4) = 24 + 0 = 24. Yep, same result.

Another example: let me take r = 5. Then p(0) = 6*5 = 30. p(4) = 6*(4 - 5) = 6*(-1) = -6. Then p(0) + p(4) = 30 - 6 = 24. Yep.

Therefore, regardless of the value of r, the sum is 24. Therefore, the answer is 24.

But wait, why does this happen? Is there a deeper reason? Maybe considering the polynomial's symmetry? Let me think.

Suppose we consider the polynomial p(x) with roots 1, 2, 3, r. Then, if we consider the function f(x) = p(x) * something... Maybe not. Alternatively, consider that p(0) is product of roots, which is 6r. And p(4) is... Well, when we plug in x=4, it's related to the value of the polynomial at 4. However, when we add p(0) and p(4), the terms with r cancel.

Alternatively, think of the polynomial as being symmetric in some way? For example, if we have roots symmetric around some point. But with roots at 1, 2, 3, and r, unless r is chosen such that there's symmetry. However, the fact that the sum p(0) + p(4) is constant regardless of r seems to be a coincidence of the specific form of the polynomial.

Alternatively, think of p(0) + p(4) as something involving the roots. But since p(0) = 6r and p(4) = 6(4 - r), adding them cancels the r term.

So maybe this is a general result? For example, if you have a monic polynomial with roots a, b, c, d, then p(0) is the product of the roots, and p(k) is something... But in this case, with the specific roots 1, 2, 3, and r, and evaluating at 0 and 4, the expression p(0) + p(4) ends up being constant.

But regardless, the math checks out with multiple examples and through two different methods (factoring and expanding). So I think the answer is indeed 24.

Therefore, I think the answer is \boxed{24}.

**Final Answer**
\boxed{24}

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

---

### Step 1: General Form of $ p(x) $

Since $ p(x) $ is a fourth-degree monic polynomial, and we know three of its roots, the general form must be:

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

where $ r $ is the unknown fourth root. This form guarantees that $ p(x) $ is monic, and all four roots are taken into account, counted with possible multiplicities.

---

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

We substitute $ x = 0 $ into the polynomial:

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

Multiply the factors:

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

So,

$$
p(0) = 6r
$$

---

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

Now substitute $ x = 4 $ into the same expression:

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

Multiply the constants:

$$
3\cdot2\cdot1 = 6,\quad 6\cdot(4 - r) = 24 - 6r
$$

So,

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

---

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

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

Simplify:

$$
6r - 6r + 24 = 24
$$

---

### Final Result

The value of $ p(0) + p(4) $ is **independent of the value of $ r $**, and always equals:

$$
\boxed{24}
$$

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 connect these.

First, maybe I should find out how daps relate to baps by using the yaps as a middle unit. That is, convert daps to yaps first, then yaps to baps. Let me write down the given equations:

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

So, if I can express both daps and baps in terms of yaps, I can find a relationship between daps and baps. Let me see. Let me try to find how many yaps are in 1 dap, and how many baps are in 1 yap, then maybe combine them?

Starting with the first equation: 4 daps = 7 yaps. So, if I divide both sides by 4, I get 1 dap = 7/4 yaps. Let me write that:

1 dap = 7/4 yaps.

Okay, so that's daps to yaps. Now, for the second equation: 5 yaps = 3 baps. So, to find how many baps are in 1 yap, I can divide both sides by 5:

1 yap = 3/5 baps.

So now, if I have 1 dap equals 7/4 yaps, and each of those yaps is equal to 3/5 baps, then I can substitute that in to find how many baps are in 1 dap. Let me do that:

1 dap = 7/4 yaps * 3/5 baps per yap.

Multiplying those fractions: (7/4) * (3/5) = (7*3)/(4*5) = 21/20 baps.

So, 1 dap equals 21/20 baps. Therefore, if I want to find how many daps equal 42 baps, I can set up a proportion or an equation.

Let me call the number of daps needed to equal 42 baps as "x". Then:

x daps = 42 baps.

But since 1 dap is 21/20 baps, then x daps would be x*(21/20) baps. Therefore:

x*(21/20) = 42.

Solving for x, I can multiply both sides by 20/21:

x = 42 * (20/21).

Calculating that: 42 divided by 21 is 2, so 2*20 is 40. Therefore, x = 40.

Wait, so 40 daps equal 42 baps? Let me check my steps again to make sure I didn't make any mistakes.

First, converting 4 daps to 7 yaps. Then converting 5 yaps to 3 baps. Then I found that 1 dap is 7/4 yaps. Then 1 yap is 3/5 baps. Then multiplying those together gives 7/4 * 3/5 = 21/20 baps per dap. So, 1 dap = 21/20 baps. Therefore, 21/20 baps per dap. So to get 42 baps, divide 42 by 21/20. Which is 42 * 20/21. 42 divided by 21 is 2, so 2*20=40. That seems right.

Alternatively, maybe I can approach this problem by converting baps to yaps first, then yaps to daps. Let's try that method to verify.

Given that 5 yaps = 3 baps, so if I want to find how many yaps are in 42 baps, I can set up a proportion. Let's solve for yaps in 42 baps.

Let me denote yaps as Y.

So, 5 yaps / 3 baps = Y / 42 baps.

Cross multiplying: 5/3 * 42 = Y.

Calculating that: 5 * 14 = 70. Wait, because 42 divided by 3 is 14, then multiplied by 5 is 70. So, Y = 70 yaps. Therefore, 42 baps = 70 yaps.

Now, since 4 daps = 7 yaps, how many daps are in 70 yaps?

Let me denote daps as D.

So, 4 daps / 7 yaps = D / 70 yaps.

Cross multiplying: 4/7 * 70 = D.

Calculating that: 4 * 10 = 40. So, D = 40 daps.

So that's the same answer as before. Therefore, 42 baps equal 40 daps.

So both methods lead me to the same answer. That gives me more confidence that it's correct. Let me check once more with another approach.

Alternatively, I can use dimensional analysis. Starting with 42 baps, convert to yaps, then to daps.

Starting with 42 baps. Using the second conversion: 5 yaps = 3 baps. So, to convert baps to yaps:

42 baps * (5 yaps / 3 baps) = 42 * 5 / 3 yaps = (42 / 3) * 5 yaps = 14 * 5 yaps = 70 yaps.

Then convert 70 yaps to daps using the first conversion: 4 daps = 7 yaps. Therefore:

70 yaps * (4 daps / 7 yaps) = 70 * 4 / 7 daps = 10 * 4 daps = 40 daps.

Same answer again. So that's three different ways of thinking through the problem, and each time I get 40 daps. So I think that's solid.

Alternatively, maybe I can set up ratios and solve algebraically. Let me try that.

Let me denote:

From 4 daps = 7 yaps, we can write the ratio of daps to yaps as 4:7.

From 5 yaps = 3 baps, the ratio of yaps to baps is 5:3.

So, combining these ratios, we can find the ratio of daps to baps.

To combine ratios, we need the common term, which is yaps. So, the ratio of daps to yaps is 4:7 and yaps to baps is 5:3. So to combine them, we need to make the number of yaps the same in both ratios.

The first ratio has 7 yaps, the second has 5 yaps. So we need to find a common multiple of 7 and 5, which is 35.

Therefore, scale the first ratio (daps:yaps) by 5 to get 20:35. Scale the second ratio (yaps:baps) by 7 to get 35:21.

Therefore, combining these, the ratio of daps:yaps:baps is 20:35:21.

Therefore, 20 daps = 35 yaps = 21 baps.

So, from this ratio, 20 daps equal 21 baps. Therefore, to find how many daps equal 42 baps, we can set up a proportion:

20 daps / 21 baps = x daps / 42 baps

Cross multiplying: 21x = 20 * 42

Therefore, x = (20 * 42)/21 = 20 * 2 = 40.

Again, same answer. So that's another method, ratios, which gives the same result.

So, after checking multiple methods—conversion factors, dimensional analysis, ratios—all leading to 40 daps equaling 42 baps. I feel confident that 40 is the correct answer.

But let me just think if there's any chance I messed up the fractions. For example, in the first method, I had 1 dap = 7/4 yaps. Then 1 yap = 3/5 baps. So multiplying 7/4 * 3/5 = 21/20. That seems correct. So 1 dap is 21/20 baps. Therefore, to get 42 baps, divide by 21/20, which is multiply by 20/21.

So, 42 * 20 / 21. Which is 2 * 20 = 40. Correct.

Alternatively, if I think in terms of units. Suppose I have 42 baps. How do I convert to daps?

We know that 5 yaps = 3 baps. So 1 bap = 5/3 yaps. Therefore, 42 baps = 42 * (5/3) yaps = 70 yaps. Then, since 4 daps = 7 yaps, then 1 yap = 4/7 daps. Therefore, 70 yaps = 70 * (4/7) daps = 10 * 4 = 40 daps. Yep, same answer.

So, all methods converge. Therefore, the answer is 40.

I think that's thorough. I can't see where I made a mistake here. Maybe I can check with numbers. Let me try plugging in the numbers.

Suppose 4 daps = 7 yaps. Let me assign numbers. Let me say 1 dap is 7/4 yaps. Let me take 4 daps = 7 yaps.

Then, 5 yaps = 3 baps, so 1 yap = 3/5 baps.

So, 7 yaps = 7*(3/5) baps = 21/5 baps. Therefore, 4 daps = 21/5 baps. Therefore, 1 dap = 21/20 baps.

So, 1 dap = 21/20 baps. Therefore, 40 daps = 40*(21/20) = 2*21 = 42 baps.

Exactly! That checks out. So, 40 daps is 42 baps. Perfect, that's exactly what we wanted.

So, all the steps are consistent. Therefore, the answer is 40.

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

To solve the problem, we are given two conversion rates:

- $ 4 \text{ daps} = 7 \text{ yaps} $
- $ 5 \text{ yaps} = 3 \text{ baps} $

Our goal is to find how many **daps** equal **42 baps**. To do this, we'll convert **baps to yaps** and then **yaps to daps**, using a common unit—**yaps**—as a bridge.

---

### Step 1: Convert baps to yaps

Given:
$$
5 \text{ yaps} = 3 \text{ baps} \Rightarrow \frac{5}{3} \text{ yaps per bap}
$$

So, for 42 baps:
$$
42 \text{ baps} \times \frac{5 \text{ yaps}}{3 \text{ baps}} = \frac{42 \cdot 5}{3} = 14 \cdot 5 = 70 \text{ yaps}
$$

---

### Step 2: Convert yaps to daps

Given:
$$
4 \text{ daps} = 7 \text{ yaps} \Rightarrow \frac{4}{7} \text{ daps per yap}
$$

So, for 70 yaps:
$$
70 \text{ yaps} \times \frac{4 \text{ daps}}{7 \text{ yaps}} = \frac{70 \cdot 4}{7} = 10 \cdot 4 = 40 \text{ daps}
$$

---

### Final Answer

$$
\boxed{40}
$$

Thus, **42 baps are equivalent 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 connect these.

First, maybe I should find out how daps relate to baps by using the yaps as a middle unit. That is, convert daps to yaps first, then yaps to baps. Let me write down the given equations:

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

So, if I can express both daps and baps in terms of yaps, I can find a relationship between daps and baps. Let me see. Let me try to find how many yaps are in 1 dap, and how many baps are in 1 yap, then maybe combine them?

Starting with the first equation: 4 daps = 7 yaps. So, if I divide both sides by 4, I get 1 dap = 7/4 yaps. Let me write that:

1 dap = 7/4 yaps.

Okay, so that's daps to yaps. Now, for the second equation: 5 yaps = 3 baps. So, to find how many baps are in 1 yap, I can divide both sides by 5:

1 yap = 3/5 baps.

So now, if I have 1 dap equals 7/4 yaps, and each of those yaps is equal to 3/5 baps, then I can substitute that in to find how many baps are in 1 dap. Let me do that:

1 dap = 7/4 yaps * 3/5 baps per yap.

Multiplying those fractions: (7/4) * (3/5) = (7*3)/(4*5) = 21/20 baps.

So, 1 dap equals 21/20 baps. Therefore, if I want to find how many daps equal 42 baps, I can set up a proportion or an equation.

Let me call the number of daps needed to equal 42 baps as "x". Then:

x daps = 42 baps.

But since 1 dap is 21/20 baps, then x daps would be x*(21/20) baps. Therefore:

x*(21/20) = 42.

Solving for x, I can multiply both sides by 20/21:

x = 42 * (20/21).

Calculating that: 42 divided by 21 is 2, so 2*20 is 40. Therefore, x = 40.

Wait, so 40 daps equal 42 baps? Let me check my steps again to make sure I didn't make any mistakes.

First, converting 4 daps to 7 yaps. Then converting 5 yaps to 3 baps. Then I found that 1 dap is 7/4 yaps. Then 1 yap is 3/5 baps. Then multiplying those together gives 7/4 * 3/5 = 21/20 baps per dap. So, 1 dap = 21/20 baps. Therefore, 21/20 baps per dap. So to get 42 baps, divide 42 by 21/20. Which is 42 * 20/21. 42 divided by 21 is 2, so 2*20=40. That seems right.

Alternatively, maybe I can approach this problem by converting baps to yaps first, then yaps to daps. Let's try that method to verify.

Given that 5 yaps = 3 baps, so if I want to find how many yaps are in 42 baps, I can set up a proportion. Let's solve for yaps in 42 baps.

Let me denote yaps as Y.

So, 5 yaps / 3 baps = Y / 42 baps.

Cross multiplying: 5/3 * 42 = Y.

Calculating that: 5 * 14 = 70. Wait, because 42 divided by 3 is 14, then multiplied by 5 is 70. So, Y = 70 yaps. Therefore, 42 baps = 70 yaps.

Now, since 4 daps = 7 yaps, how many daps are in 70 yaps?

Let me denote daps as D.

So, 4 daps / 7 yaps = D / 70 yaps.

Cross multiplying: 4/7 * 70 = D.

Calculating that: 4 * 10 = 40. So, D = 40 daps.

So that's the same answer as before. Therefore, 42 baps equal 40 daps.

So both methods lead me to the same answer. That gives me more confidence that it's correct. Let me check once more with another approach.

Alternatively, I can use dimensional analysis. Starting with 42 baps, convert to yaps, then to daps.

Starting with 42 baps. Using the second conversion: 5 yaps = 3 baps. So, to convert baps to yaps:

42 baps * (5 yaps / 3 baps) = 42 * 5 / 3 yaps = (42 / 3) * 5 yaps = 14 * 5 yaps = 70 yaps.

Then convert 70 yaps to daps using the first conversion: 4 daps = 7 yaps. Therefore:

70 yaps * (4 daps / 7 yaps) = 70 * 4 / 7 daps = 10 * 4 daps = 40 daps.

Same answer again. So that's three different ways of thinking through the problem, and each time I get 40 daps. So I think that's solid.

Alternatively, maybe I can set up ratios and solve algebraically. Let me try that.

Let me denote:

From 4 daps = 7 yaps, we can write the ratio of daps to yaps as 4:7.

From 5 yaps = 3 baps, the ratio of yaps to baps is 5:3.

So, combining these ratios, we can find the ratio of daps to baps.

To combine ratios, we need the common term, which is yaps. So, the ratio of daps to yaps is 4:7 and yaps to baps is 5:3. So to combine them, we need to make the number of yaps the same in both ratios.

The first ratio has 7 yaps, the second has 5 yaps. So we need to find a common multiple of 7 and 5, which is 35.

Therefore, scale the first ratio (daps:yaps) by 5 to get 20:35. Scale the second ratio (yaps:baps) by 7 to get 35:21.

Therefore, combining these, the ratio of daps:yaps:baps is 20:35:21.

Therefore, 20 daps = 35 yaps = 21 baps.

So, from this ratio, 20 daps equal 21 baps. Therefore, to find how many daps equal 42 baps, we can set up a proportion:

20 daps / 21 baps = x daps / 42 baps

Cross multiplying: 21x = 20 * 42

Therefore, x = (20 * 42)/21 = 20 * 2 = 40.

Again, same answer. So that's another method, ratios, which gives the same result.

So, after checking multiple methods—conversion factors, dimensional analysis, ratios—all leading to 40 daps equaling 42 baps. I feel confident that 40 is the correct answer.

But let me just think if there's any chance I messed up the fractions. For example, in the first method, I had 1 dap = 7/4 yaps. Then 1 yap = 3/5 baps. So multiplying 7/4 * 3/5 = 21/20. That seems correct. So 1 dap is 21/20 baps. Therefore, to get 42 baps, divide by 21/20, which is multiply by 20/21.

So, 42 * 20 / 21. Which is 2 * 20 = 40. Correct.

Alternatively, if I think in terms of units. Suppose I have 42 baps. How do I convert to daps?

We know that 5 yaps = 3 baps. So 1 bap = 5/3 yaps. Therefore, 42 baps = 42 * (5/3) yaps = 70 yaps. Then, since 4 daps = 7 yaps, then 1 yap = 4/7 daps. Therefore, 70 yaps = 70 * (4/7) daps = 10 * 4 = 40 daps. Yep, same answer.

So, all methods converge. Therefore, the answer is 40.

I think that's thorough. I can't see where I made a mistake here. Maybe I can check with numbers. Let me try plugging in the numbers.

Suppose 4 daps = 7 yaps. Let me assign numbers. Let me say 1 dap is 7/4 yaps. Let me take 4 daps = 7 yaps.

Then, 5 yaps = 3 baps, so 1 yap = 3/5 baps.

So, 7 yaps = 7*(3/5) baps = 21/5 baps. Therefore, 4 daps = 21/5 baps. Therefore, 1 dap = 21/20 baps.

So, 1 dap = 21/20 baps. Therefore, 40 daps = 40*(21/20) = 2*21 = 42 baps.

Exactly! That checks out. So, 40 daps is 42 baps. Perfect, that's exactly what we wanted.

So, all the steps are consistent. Therefore, the answer is 40.

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

To solve the problem, we are given two conversion rates:

- $ 4 \text{ daps} = 7 \text{ yaps} $
- $ 5 \text{ yaps} = 3 \text{ baps} $

Our goal is to find how many **daps** equal **42 baps**. To do this, we'll convert **baps to yaps** and then **yaps to daps**, using a common unit—**yaps**—as a bridge.

---

### Step 1: Convert baps to yaps

Given:
$$
5 \text{ yaps} = 3 \text{ baps} \Rightarrow \frac{5}{3} \text{ yaps per bap}
$$

So, for 42 baps:
$$
42 \text{ baps} \times \frac{5 \text{ yaps}}{3 \text{ baps}} = \frac{42 \cdot 5}{3} = 14 \cdot 5 = 70 \text{ yaps}
$$

---

### Step 2: Convert yaps to daps

Given:
$$
4 \text{ daps} = 7 \text{ yaps} \Rightarrow \frac{4}{7} \text{ daps per yap}
$$

So, for 70 yaps:
$$
70 \text{ yaps} \times \frac{4 \text{ daps}}{7 \text{ yaps}} = \frac{70 \cdot 4}{7} = 10 \cdot 4 = 40 \text{ daps}
$$

---

### Final Answer

$$
\boxed{40}
$$

Thus, **42 baps are equivalent 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.