MMOD29 - CALCULATE POW(2004,X) MOD 29
English | Vietnamese |
Consider a positive integer X, and let S be the sum of all positive integer divisors of 2004X. Your job is to determine S modulo 29 (the remainder of the division of S by 29).
Take X = 1 for an example. The positive integer divisors of 20041 are 1, 2, 3, 4, 6, 12, 167, 334, 501, 668, 1002 and 2004. Therefore S = 4704 and S modulo 29 is equal to 6.
Input
The input consists of several test cases. Each test case contains a line with the integer X (1 <= X <= 10000000). A test case of X = 0 indicates the end of input, and should not be processed.
Output
For each test case, in a separate line, please output the result of S modulo 29.
Sample
Input: 1 10000 0 Output: 6 10
Added by: | psetter |
Date: | 2009-02-21 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | Peiking 2004 |