IMO 1979 SL 7

Origin: FRG

Problem

Given that 1 −1 2 + 1 3 −1 4 + \cdot \cdot \cdot − 1318 + 1319 = p q , where p and q are natural numbers having no common factor, prove that p is divisible by 1979.

Solution

We denote the sum mentioned above by S. We have the following equali- ties:

S = 1 −1 2 + 1 3 −1 4 + \cdot \cdot \cdot − 1318 + 1319 = 1 + 1 2 + \cdot \cdot \cdot + 1319 −2 1 2 + 1 4 + \cdot \cdot \cdot + 1318  = 1 + 1 2 + \cdot \cdot \cdot + 1319 −  1 + 1 2 + \cdot \cdot \cdot + 

660 + 661 + \cdot \cdot \cdot + 1319

 i=660 i + 1979 −i =  i=660 1979 i \cdot (1979 −i) Since no term in the sum contains a denominator divisible by 1979 (1979 is a prime number), it follows that when S is represented as p/q the numerator p will have to be divisible by 1979.