If x = 1! + 2! + 3! +....... + n!, what is the remainder when x is divided by 25?
1. n>8
2. n<10
Will post OA in a while....
1. n>8
2. n<10
Will post OA in a while....
imo A
I haven't solved it but here is my approach
Statement I
n>8
10! and beyond, all numbers are divisible by 25.
10! = 10 *9*8*7*6*5*4*3*2*1
or 10! = 5*2*9*8*7*6*5*4*3*2*1
and 11! = 11*10!
and so on..
so basically if we find the remainder for the sum of the series 1! + 2! +3!...9!
that'll be the remainder for the entire series.
This will make statement I sufficient
Statement II
n<10
now, the remainder may vary because n could be 1, 2, 3, 4, 5, 6, 7, 8 or 9
lets say n = 4
1! + 2+ 3! + 4! = 1+2+6+24 = 33
33/25 ------ remainder = 8
now n=6
1! +2!+3!+4!+5!+6!
1+2+6+24+120+720 = 840+33 = 873
873/25---------remainder = 23
Hence, Statement II is insufficient
