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 Numerator Divisibility Nuance (Posted on 2011-01-22)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution | Comment 3 of 4 |

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.

 Posted by Brian Smith on 2011-01-23 14:55:04

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