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.

There are 25 primes between 1 and 100 inclusive. The number of prime factors of the product of 1! x 2! x 3! x ... x 98! x 99! x 100! is as follows:

 2 : 4731
 3 : 2328
 5 : 1124
 7 :  734
11 :  414
13 :  343
17 :  250
19 :  220
23 :  174
29 :  129
31 :  117
37 :   91
41 :   79
43 :   73
47 :   61
53 :   48
59 :   42
61 :   40
67 :   34
71 :   30
73 :   28
79 :   22
83 :   18
89 :   12
97 :    4

As there are 1124 factors of 5 and at least that many factors of 2, there are 1124 trailing zeros; and, thus, T equals 1124.

The remainder of 1124 divided by 1000 is 124.



Though not asked for, the T^th+1 digit from the right can be determined as follows... Subtract 1124 from the number of factors of both 2 and 5 leaving 3607 factors of 2 and zero of 5. The counts of the ones-digits of each remaining factors are as follows:
2 : 3607
3 : 2328
7 : 1174
9 :  425

The ones-digit of 2ⁿ has a cyclic period of 4 {2,4,8,6}.
3607 modulo 4 is 3, therefore, the ones-digit of 2³⁶⁰⁷ is 8.
The ones-digit of 3ⁿ has a cyclic period of 4 {3,9,7,1}.
2328 modulo 4 is 0, therefore, the ones-digit of 3²³²⁸ is 1.
The ones-digit of 7ⁿ has a cyclic period of 4 {7,9,3,1}.
1174 modulo 4 is 2, therefore, the ones-digit of 7¹¹⁷⁴ is 9.
The ones-digit of 9ⁿ has a cyclic period of 2 {9,1}.
425 modulo 2 is 1, therefore, the ones-digit of 9⁴²⁵ is 9.

Since, 8 x 1 x 9 x 9 = 648, the T^th+1 digit from the right is an 8.

Edited on November 25, 2008, 4:08 pm
Corrected the errors noted by Charlie.

Edited on November 25, 2008, 11:10 pm

Posted by Dej Mar on 2008-11-25 15:32:56