IMO problem: 1979-1 (Easy)

[sirio11]

Let p/q = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ..... + 1/1319

with (p,q)=1 (This is the fraction is irreducible)

prove that 1979 divides p

[blah_blah]

important: 1979 is a prime number! /images/graemlins/smile.gif

[jay_shark]

This is pretty neat and I remember solving this exact problem several years ago . It happens to be the year of my birthyear /images/graemlins/smile.gif

The trick is to re-write the sum as :

p/q = 1 + 1/2 + 1/3 + 1/4 + 1/5 +...+1/1319 -2*1/2 - 2*1/4 - 2*1/6 ...-2*1/1318.

The negative terms can be written as -2*(1/2+1/4+1/6+...+1/1318) or -1*(1+1/2+1/3+1/4+...+1/659)

p/q = 1/660 + 1/661 + 1/662 +...+ 1/1319

Notice that (1/660 + 1/1319)+ (1/661+1/1318)+...
=1979/(660*1319)+ 1979/(661*1318)+...
=1979*[1/(660*1319)+1/(661*1318)+1/(662*1317)+...]

Since all the denominators in the square brackets are less than 1979 and that 1979 is prime , we have that p is divisible by 1979 .

[Enrique]

[ QUOTE ]
Let p/q = 1 - 1/2 + 1/3 - 1/4 + 1/5 - ..... + 1/1319

with (p,q)=1 (This is the fraction is irreducible)

prove that 1979 divides p

[/ QUOTE ]

it seems redundant to say (p,q) = 1.
If (p,q) = k, then you can prove 1979 divides p/k, hence 1979 divides p.
So the divisibility property happens regardless of grabbing the irreducible fraction.