## tricky question

Q1. An old man has Rs (1! + 2! + 3! + ...+ 50!), all of which he wants to divide equally (without fractions) among his n children. Then, n may be (a)5 (b)7 (c)9 (d)11 wat cud b da ansr......guyzzz

## The answer is 9.Any number

The answer is 9. Any number more than 6!(it also) is always divisible by 9 because it contains at least two 3's and sum of first 5 number factorials is divisible by 9 1! + 2! + 3! + 4! + 5! = 153 which is divisible by 9.