Search found 1 match
n - odd => n=2k+1, k=0,1,2,...
n^2-1= (n+1)(n-1) = (2k+2)*2k= 4k*(k+1)
4k*(k+1)/8=1/2*k*(k+1) => k or k+1 is even => remainder (n^2-1)/8 is 0
- by vorand
Fri Nov 12, 2010 11:49 am- Forum: Data Sufficiency
- Topic: Remainder Prob
- Replies: 74
- Views: 40047