Each of the last T digits in the decimal representation of the product of 1!*2!*3!.....99!*100! is zero, but the (T+1)^th digit from the right is nonzero.

Determine the remainder when T is divided by 1000.

Note: Try to derive a non computer-assisted method, although computer program/spreadsheet solutions are welcome.

Let  NOZ(t) denote the number of consecutive zeroes
 in the rightmost part of  t.

Clearly NOZ(T!)= NOZ((T-1)!)  unless T is a multiple of 5.      èrule A

If T is a multiple of 5^k (k>0) then NOZ(T!)= NOZ((T-1)!)+k.  è rule B

                                                                                              The first factorial ending with  one zero is 5!
  So are 6!,7!  8! and  9!- rule A--  but NOZ(10!)=2, NOZ(15!)=3, NOZ(25!)=6  etc! Therefore multiplying all factorials  between 5! and 99! will get us five times the sum of all the integers between 1 and 22 inclusive , omitting 5, 11 ,  and 17 è rule B.

So NOZ(1!*2!*3!*4!*…..98!*99!)= 5*(1/2*(1+22)*22-5-1-17)=5*( 253-33)= 1100
Adding  NOZ(100)=24 we get 1124.

Since the puzzle requests the answer mod 1000 =è we end up with   124.<o:p></o:p>

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Edited on November 28, 2008, 12:47 am

Posted by Ady TZIDON on 2008-11-27 18:19:08