Question

Find the maximum value of n such that 50! is perfectly divisible by ${12600}^n$.

• A. 7
• B. 6
• C. 8
• D. None of these

Answer 1

The correct option is B 6

$12600=7\times 3^2\times 2^3\times 5^2$
The value of 'n' would depend on which of number of 7s and number of $5^{2}s$ is lower in 50!.
Number of 7's in 50! = 8.
Note here that if we check for 7's we do not need to check for $3^{2}s$ as there would be at least two 3's before a 7 comes in every factorial's value. Similarly, there would always be at least three 2's before a 7 comes in any factorial's value. Thus, the number of $3^{2}s$ and the number of $2^{3}s$ can never be lower than the number of 7s in any factorial's value.
Number of 5s in 50! = 10 + 2 =12. Hence, the number of $5^{2}s$ in 50! = $\left[\frac{12}{2}\right]=6$
6 will be the answer as the number of $5^{2}s$ is lower than the number of 7's.
Option (b) is correct.