How to calculate sum of this series modulo m fast?

Question

so I came across this problem where I need to calculate this:

1^k+(1+p)^k+(1+2*p)^k+.....+(1+n*p)^k % p

Where p is prime and k is some number strictly less than p.

p is less than 500 and n*p could range upto 10⁹

The only solution I could think is iterate from first to last term and calculate the modulo using exponentiation but that would be too costly I am looking for a faster algorithm.

Is it possible to do it faster?

@JohnColeman ohh my! Not exactly a homework, but looks like I didn't think enough. — anker, Jul 25 '17 at 16:35
@JohnColeman looks like it, can I accept a comment as an answer? — anker, Jul 25 '17 at 16:37

John Coleman · Accepted Answer · 2017-07-25T20:32:22.913

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For any integer m, (1+m*p)^k % p == 1.

Thus, computing

(1^k + (1+2*p)^k  + (1+3*p)^k + ... + (1+n*p)^k )% p

is the same as computing

(1 + 1 + 1 ... + 1) % p

Where there are n + 1 terms in the parentheses.

The answer is thus (n + 1)%p.

edited Jul 25 '17 at 20:32

answered Jul 25 '17 at 16:50

John Coleman

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I can't imagine someone having a faster algorithm than that! :P – SirGuy Jul 25 '17 at 16:59
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There are n+1 terms. – Jul 25 '17 at 20:25

How to calculate sum of this series modulo m fast?

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