Question

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

Hint:

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

Answer 1

Correct Answer: 14

Step 1: Prime factorization of 40.
\[ 40 = 2^3 \times 5 \] \[ 40^n = 2^{3n} \times 5^n \] Step 2: Highest power of 2 in $60!$.
\[ \left\lfloor \frac{60}{2} \right\rfloor + \left\lfloor \frac{60}{4} \right\rfloor + \left\lfloor \frac{60}{8} \right\rfloor + \left\lfloor \frac{60}{16} \right\rfloor + \left\lfloor \frac{60}{32} \right\rfloor \] \[ = 30 + 15 + 7 + 3 + 1 = 56 \] Step 3: Highest power of 5 in $60!$.
\[ \left\lfloor \frac{60}{5} \right\rfloor + \left\lfloor \frac{60}{25} \right\rfloor \] \[ = 12 + 2 = 14 \] Step 4: Determine limiting factor.
From $2^{3n} \leq 2^{56}$, we get $n \leq 18$.
From $5^n \leq 5^{14}$, we get $n \leq 14$.
Step 5: Final conclusion.
Hence, the maximum value of $n$ is 14.