Question

Find the maximum value of n such that $42\times 57\times 92\times 91\times 52\times 62\times 63\times 64\times 65\times 66\times 67$ is perfectly divisible by ${42}^n$.

• A. 4
• B. 3
• C. 5
• D. 6

Answer 1

The correct option is B 3

42 = $7\times 3\times 2$
Also, 57 = $3\times 19$
92 = $2\times 2\times 23$
91 = $7\times 13$
52 = $2\times 2\times 13$
62 = $2\times 31$
63 = $3\times 3\times 7$
64 = $2\times 2\times 2\times 2\times 2\times 2$
65 = $5\times 13$
66 = $2\times 3\times 11$
67 = $1\times 67$
In the expression given, there are three 7's and more than three 2's and 3's. Thus, Option (b) is correct.