We are given the task to find the largest value of \(n\) for which \(40^n\) divides \(60!\).

First, let's express 40 in terms of its prime factors:

\(40 = 2^3 \times 5^1\)

Therefore, \(40^n = (2^3 \times 5^1)^n = 2^{3n} \times 5^n\).

To determine the maximum value of \(n\) such that \(2^{3n} \times 5^n\) divides \(60!\), we first need to find the number of times 2 and 5 appear in the prime factorization of \(60!\).

The formula to find the exponent of a prime \(p\) in \(n!\) is:

\(\text{Exponent of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots\)

Let's apply this to primes 2 and 5:

Calculating the power of 2 in \(60!\):

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

Total power of 2 = \(30 + 15 + 7 + 3 + 1 = 56\)

Calculating the power of 5 in \(60!\):

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

Total power of 5 = \(12 + 2 = 14\)

Now, set up the equations for division:

• \(3n \leq 56\) (for power of 2)
• \(n \leq 14\) (for power of 5)

The limiting factor is the power of 5, as \(n \leq 14\) while \(3n \leq 56\) allows for \(n \leq 18.67\).

Thus, the largest integer \(n\) such that \(40^n\) divides \(60!\) is \(14\).

Therefore, the correct answer is \(14\).