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

ok, maybe I did solve it in the end

Yes that looks like the correct solution Rohan. Looks like you did solve it!

OA is A

Rohan is spot on...

The highest power of 10 in 10! is 2 => 10! ends with 00
Same with 11!, 12!, 13!, 14!..

The highest power of 10 in 15! = 3 => 15! ends with 000

The last two digits of 15! is 00

All n!; for n > 10 ends with 00

1!+2!+...+9! gives a certain remainder (which is 13 FYI)...when this sum is added with 10! or higher factorials, the last 2 digits remain 13...

Thus, the remainder is always 13...

rohangupta83 - Beautiful thinking!

Thanks guys for nice explanations.
Based on your explanation, should stmt 1 be n > 9 ?? instead of n > 8
Then I think A would be correct.

thanks
-V

Oops!!!!! I found a flaw in my reply.
Indeed u guys are correct.

thanks
-V