Question

The largest value of $n$, for which $40^n$ divides $60!$, is

• A. 13
• B. 11
• C. 14
• D. 12

Hint:

For divisibility of $a^n$ into $N!$, always compare prime powers of $a^n$ with those in $N!$ and take the minimum bound.

Answer 1

The Correct Option is C

To solve the problem of determining the largest value of \(n\), such that \(40^n\) divides \(60!\), let's start by finding the prime factorization of 40. We have:

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

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

To find the largest \(n\) for which \(40^n\) divides \(60!\), we need to calculate how many times 2 and 5 appear in the prime factorization of \(60!\). We use the formula for the highest power of a prime \(p\) dividing \(n!\):

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

Calculate for 2 in \(60!\):

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

Therefore, the power of 2 in \(60!\) is \(30 + 15 + 7 + 3 + 1 = 56\).

Calculate for 5 in \(60!\):

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

Therefore, the power of 5 in \(60!\) is \(12 + 2 = 14\).

Now, for \(40^n = 2^{3n} \times 5^n\) to divide \(60!\), we have the conditions:

• \(3n \leq 56\)
• \(n \leq 14\)

From the equation \(3n \leq 56\), we get \(n \leq \lfloor \frac{56}{3} \rfloor = 18\). However, \(n \leq 14\) due to the limitation of the factor 5.

The largest value of \(n\) is therefore \(14\), since it satisfies both conditions.

Hence, the correct answer is 14.