Question

1(1!)+2(2!)+3(3!)+--------+19(19!)

find the remainder of the sum of this series when it is divided by 64 ?

Answer 1

Notice first that 8! is divisible by 64, as 8 = 1.2.3.4.5.6.7.8 and 2.4.8 = 64;
we write 8! | 64

By the same argument j! | 64 for all j >= 8. Thus in the sum of the question,
S = 1.1! + 2.2! + ... + 19.19!
we can disregard all terms after 7.7! as all of them will have remainder 0 when divided by 64.

Now term by term,

1.1! = 1, which when divided by 64 has remainder r = 1
2.2! = 4 , r = 4
3.3! = 18, r = 18
4.4! = 96, r = 32
5.5! = 600, r = 24
6.6! = 4320, r = 32
7.7! = 35,280, r = 16

The sum of those remainders is 127, which has remainder 63.

This suggests there might be some further simplification of these last 7 terms, but I can't see it off the top of my head.

Answer 2

very accuretly
u r genious........u do better work