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

Step 1: Prime factorization of the base. 
We write \[ 40 = 2^3 \times 5 \] Hence, \[ 40^n = 2^{3n} \times 5^n \] Step 2: Find powers of 2 and 5 in $60!$. 
Power of 2 in $60!$ is \[ \left\lfloor \frac{60}{2} \right\rfloor + \left\lfloor \frac{60}{4} \right\rfloor + \left\lfloor \frac{60}{8} \right\rfloor + \left\lfloor \frac{60}{16} \right\rfloor + \left\lfloor \frac{60}{32} \right\rfloor \] \[ = 30 + 15 + 7 + 3 + 1 = 56 \] Power of 5 in $60!$ is \[ \left\lfloor \frac{60}{5} \right\rfloor + \left\lfloor \frac{60}{25} \right\rfloor = 12 + 2 = 14 \] Step 3: Apply divisibility condition. 
For $40^n$ to divide $60!$, we must have \[ 3n \le 56 \quad \text{and} \quad n \le 14 \] From the first condition, \[ n \le \frac{56}{3} \approx 18 \] Step 4: Determine the limiting value. 
The limiting condition is $n \le 14$. 
Step 5: Final conclusion. 
Hence, the largest possible value of $n$ is 14.