Question

Find the maximum value of n such that 77! is perfectly divisible by ${720}^n$.

• A. 35
• B. 18
• C. 17
• D. 36

Answer 1

The correct option is C 17

$720=2^4\times 5^1\times 3^2$
In 77! there would be 38 + 19 + 9 + 4 + 2 + 1 = 73 twos $\to$ hence $\left[\frac{73}{4}\right]=182^{4}s$
In 77! there would be 25 + 8 + 2 = 35 threes $\to$ hence $\left[\frac{35}{2}\right]=173^{2}s$
In 77! there would be 15 + 3 = 18 fives
Since 17 is the least of these values, option (c) is correct.