Remainder When N divided by 2^M

scanf("%d", &T);
while(T--){
   scanf("%d,%d", &N,&M);
   printf("%d",N&~(~0 << M));
}

Uh, guys, in pretty much any language, it would be

N & (1 << M  - 1)

Golfscript - 16 chars

~]1>2/{~2\?(&n}/

For a single pair or numbers this would sufficient

~2\?(&

By the way, the / does not stand for division here :)