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

Q. The largest value of \(n\), for which \(40^n\) divides \(60!\), is

(A) 14

(B) 13

(D) 12

Correct Answer: 14

Explanation

First, write \(40\) in its prime factorised form.

\[ 40 = 2^3 \times 5 \]

Therefore,

\[ 40^n = 2^{3n} \times 5^n \]

For \(40^n\) to divide \(60!\), the power of \(2\) in \(60!\) must be at least \(3n\) and the power of \(5\) must be at least \(n\).

Now calculate the power of \(2\) in \(60!\).

\[ \left\lfloor \frac{60}{2} \right\rfloor + \left\lfloor \frac{60}{4} \right\rfloor + \left\lfloor \frac{60}{8} \right\rfloor + \left\lfloor \frac{60}{16} \right\rfloor + \left\lfloor \frac{60}{32} \right\rfloor \]

\[ = 30 + 15 + 7 + 3 + 1 = 56 \]

Hence, power of \(2\) in \(60!\) is \(56\).

Now calculate the power of \(5\) in \(60!\).

\[ \left\lfloor \frac{60}{5} \right\rfloor + \left\lfloor \frac{60}{25} \right\rfloor \]

\[ = 12 + 2 = 14 \]

Thus, \(n\) must satisfy

\[ 3n \le 56 \quad \text{and} \quad n \le 14 \]

From \(3n \le 56\),

\[ n \le \frac{56}{3} \approx 18 \]

Combining both conditions, the largest possible integer value of \(n\) is

\[ \boxed{14} \]

Therefore, the correct answer is 14.

Explanation

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