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.

Well, the key to finding T is to count how many powers of 10 (ie,2*5) are in the factorial.  Since there are at lot more 2's than 5's, all we need to do is count the powers of 5.

The given factorial is equivalent to:  100*99^2*98^3*97^4 ...2^99

For purposes of counting 5's, all we need are the factors that include a 5:   100*95^6*90^11*...5^96. 

factor x contributes (101-x) 5's, unless it is a multiple of 25, in which case it contributes 2*(101-x) 5's.  My tally is

100    2 <-- multiple of 25
95      6
90    11
85    16
80    21
75    52 <-- multiple of 25
70    31
65    36
60    41
55    46
50  102 <-- multiple of 25
45    56
40    61
35    66
30    71
25  152 <-- multiple of 25
20    81
15    86
10    91
5      96
      ----
T = 1124, so the requested answer is 124

Posted by Steve Herman on 2008-11-25 14:47:17