+41 If a and b are integers, such that a does not equal 0 and b doe not equal 0 and and the quares of a and b have at most two digits, what is the greatest possible difference between the squares of a and b
+41 the solution is
so the max a or b's square is 81(9x9)
and the least is 1 (1x1)
81-1=80
