N! is completely divisible by `13^(52)`. What is sum of the digits of the smallest such number N?

Options

A. 11
B. 15
C. 16
D. 19

Answer

To solve the problem of finding the smallest number \( N \) such that \( N! \) is completely divisible by \( 13^{52} \), we can follow these steps:

### Step 1: Understand the problem
We need to find the smallest \( N \) such that \( N! \) has at least 52 factors of 13. This means we need to determine how many times 13 appears in the prime factorization of \( N! \).

### Step 2: Use the formula for the number of factors of a prime in a factorial
The number of times a prime \( p \) divides \( N! \) can be calculated using the formula:
\[ \sum_{k=1}^{\infty} \left\lfloor \frac{N}{p^k} \right\rfloor \]
For our case, \( p = 13 \) and we want this sum to be at least 52.

### Step 3: Calculate for different values of \( N \)
We will calculate the sum for different values of \( N \) until we find the smallest \( N \) such that the sum is at least 52.

- **For \( N = 600 \)**:
\[ \left\lfloor \frac{600}{13} \right\rfloor + \left\lfloor \frac{600}{13^2} \right\rfloor = \left\lfloor \frac{600}{13} \right\rfloor + \left\lfloor \frac{600}{169} \right\rfloor = 46 + 3 = 49 \]

- **For \( N = 650 \)**:
\[ \left\lfloor \frac{650}{13} \right\rfloor + \left\lfloor \frac{650}{13^2} \right\rfloor = \left\lfloor \frac{650}{13} \right\rfloor + \left\lfloor \frac{650}{169} \right\rfloor = 50 + 3 = 53 \]

### Step 4: Determine the range for \( N \)
From the calculations:
- \( N = 600 \) gives us 49 factors of 13.
- \( N = 650 \) gives us 53 factors of 13.

Thus, \( N \) must be between 600 and 650.

### Step 5: Find the smallest \( N \) with exactly 52 factors of 13
We need to check values between 600 and 650 to find the smallest \( N \) that gives us exactly 52 factors of 13.

- **For \( N = 637 \)**:
\[ \left\lfloor \frac{637}{13} \right\rfloor + \left\lfloor \frac{637}{13^2} \right\rfloor = \left\lfloor \frac{637}{13} \right\rfloor + \left\lfloor \frac{637}{169} \right\rfloor = 49 + 3 = 52 \]

### Step 6: Conclusion
The smallest \( N \) such that \( N! \) is divisible by \( 13^{52} \) is \( N = 637 \).

### Step 7: Calculate the sum of the digits of \( N \)
Now we calculate the sum of the digits of 637:
\[ 6 + 3 + 7 = 16 \]

Thus, the sum of the digits of the smallest such number \( N \) is **16**.