Solution

Question: Find the remainder of (4*4!+ 5*5! + ... + 20*20!)/64??? Solution: Note that 64 = 2^6 and all numbers from 8! onwards are divisible by (2)^6, that is, by 64. Since (2)^6 is a factor of N! for all values of N greater than or equal to 8. So basically we have to find the remainder when (4*4! + 5*5! + 6*6! + 7*7!) is divided by 64. 4*4! = 4*24 = 96 This when divided by 64, the remainder is 32. 5*5! = 5*120 = 600 This when divided by 64, the remainder is 24. 6*6! = 6*720 = 4320 This when divided by 64, the remainder is 32. 7*7! = 7*5040 = 35280 This when divided by 64, the remainder is 18. So the sum of the remainders is: 32 + 24 + 32 + 18 Now we can remove the sum of 32 + 32 as that is perfectly divisible by 64. Hence the remainder is 24 + 18 = 42. That is, the remainder when (4*4!+5*5!+....+20*20!) is divided by 64 is 42. Thank You. Ravi Raja

Check out the

Check out the comment http://www.cat4mba.com/node/925#comment-877 You could have find it with simple search on this site

solution

(n) x n! = (n+1-1) n! = (n+1) n! x (n) n! = n+1! – n!

So 4x4!+5x5!+……………+20x20!
= 5!-4!+6!-5!+…………….+21!-20!

All terms will cancel and it will reduce to
21! - 4!
= 21!-24

21! Contains 2x4x8 i.e. 64 hence divisible by it.

Thus remainder will be for -24
The number 21! is divisible by 64 and 21!-24 means 24 less than number divisible by 64, so it needs 40 more to be exactly divisible by 24.

So remainder is 64-24 i.e. 40

i thnk ur method is rt...bt

i thnk ur method is rt...bt ur ans is nt correct,,its nt even in the options...the options r:- a)12 b)24 c)36 d)60 and i dnt knw the correct ans...!!! Try again !!!!

The answer is 40.

The answer is 40. Check out http://www.cat4mba.com/node/925#comment-496 for a very simple and easy explanation