2022/050) Find n such that $n! = 3!5!7!$

 clearly n > 7.

now if t ia factorial then 3!5! must be product of sum k sucesssive numbers from 8

3!5! = 6 * 120 = 720 = 8 * 90 = 8 * 9 * 10

so n! = 10 * 9 * 8 * 7! = 10! 

or n = 10