1i am getting p = 3, 167 (odd prime factors of 2004)
find all odd prime numbers p which divide
1p-1+2p-1+3p-1+.....................2004p-1
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4 Answers
1ap-1 ≡ 1 (mod p) where p is co-prime to a
here a varies from 1 to 2004, all of them will give remainder 1 except for a = p, 2p, 3p, ......, [2004/p] p which will give remainder 0
(here [.] is greatest integer function)
so final remainder = 2004 - [2004/p]
this should be divisible by p for which p should be odd prime factor of 2004
