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 onesdigits of each remaining factors are as follows:
2 : 3607
3 : 2328
7 : 1174
9 : 425
The onesdigit of 2
^{n} has a cyclic period of 4 {2,4,8,6}.
3607 modulo 4 is 3, therefore, the onesdigit of 2
^{3607} is 8.
The onesdigit of 3
^{n} has a cyclic period of 4 {3,9,7,1}.
2328 modulo 4 is 0, therefore, the onesdigit of 3
^{2328} is 1.
The onesdigit of 7
^{n} has a cyclic period of 4 {7,9,3,1}.
1174 modulo 4 is 2, therefore, the onesdigit of 7
^{1174} is 9.
The onesdigit of 9
^{n} has a cyclic period of 2 {9,1}.
425 modulo 2 is 1, therefore, the onesdigit of 9
^{425} 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 20081125 15:32:56 