Q. If 15X14X13x…….X3X2X1=3^mXn
where m and n are positive integers, then what is the maximum value of m?

Explanation : 
First, let’s compute the prime factorization of 15!:
15! = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
Now, let’s break down each of these factors into their prime factors:
15! = (3×5) × (2×7) × 13 × (2^2×3) × 11 × (2×5) × (3^2) × (2^3) × 7 × (2×3) × 5 × (2^2) × 3 × 2
15! = 2^11 × 3^6 × 5^3 × 7^2 × 11 × 13

We want to express this as the product of two integers, 3m and n. Since we want to maximize m, we should try to include as many factors of 3 in 3m as possible. There are 6 factors of 3 in the prime factorization of 15!, so the maximum value of m is 6.

Therefore, the maximum value of m is 6 (option b).