Each of p and q is a positive integer, with p and q being relatively prime, such that:
p/q = 1 - 1/2 + 1/3 - 1/4 + ...... - 1/1318 + 1/1319
Prove that p is divisible by 1979.
First, rewrite each negative term -1/(2n) into 1/(2n) - 1/n. This makes the sum look like 1 + (1/2 - 1) + 1/3 + (1/4 - 1/2) + ... + (1/1318 - 1/659) + 1/1319
The terms then can be rearranged as (1 - 1 + 1/2 - 1/2 + 1/3 - 1/3 + ... + 1/659 - 1/659) + (1/660 + 1/661 + ... + 1/1319)
The first sum is 0. Then rearrange the second sum into (1/660 + 1/1319) + (1/661 + 1/1318) + ... + (1/1649 + 1/1650).
Add each pair of terms in parenthesis to yield 1979/(660*1319) + 1979/(661*1318) + ... + 1979/(1649*1650). Note that each numerator is 1979.
Finding a common denominator for all the terms will not change the fact that all the individual numerators will all be multiples of 1979. Therefore the numerator of the sum must be a multiple of 1979.
Solution
