Question

Maximum value of $n$ for which $40^n$ divides $60!$ is

• A. 10
• B. 14
• C. 20
• D. 27

Hint:

When checking divisibility of factorials, always compare prime powers separately and choose the minimum.

Answer 1

The Correct Option is B

To determine the largest value of \(n\) for which \(40^n\) divides \(60!\), we need to follow these steps:

Compute the prime factorization of 40:  
\(40 = 2^3 \times 5^1\).

From the factorization, we need at least 3 factors of 2 and 1 factor of 5 per \(40\) in \(60!\).

Using the formula for finding the power of a prime \(p\) in \(n!\):

\(v_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots\)

Calculate the number of factors of 2 in \(60!\):

• \(\left\lfloor \frac{60}{2} \right\rfloor = 30\)
• \(\left\lfloor \frac{60}{4} \right\rfloor = 15\)
• \(\left\lfloor \frac{60}{8} \right\rfloor = 7\)
• \(\left\lfloor \frac{60}{16} \right\rfloor = 3\)
• \(\left\lfloor \frac{60}{32} \right\rfloor = 1\)

Calculate the number of factors of 5 in \(60!\):

• \(\left\lfloor \frac{60}{5} \right\rfloor = 12\)
• \(\left\lfloor \frac{60}{25} \right\rfloor = 2\)

Now, determine how many complete sets of factors of 40 exist in \(60!\):

We need \(3\) factors of 2 and \(1\) factor of 5 per \(40\):

• Factors of 2 needed for \(40^n\) is \(3n \leq 56\), thus \(n \leq \frac{56}{3} \approx 18.67\)
• Factors of 5 needed for \(40^n\) is \(n \leq 14\)

The limiting factor is the number of 5s, hence the maximum value of \(n\) for which \(40^n\) divides \(60!\) is \(14\).

The correct answer is therefore 14.