It helps to have:
N = n0 + 10n1 + 100n2 + ...
M = m0 + 10m1 + 100m2 + ...
where the undercase ms and ns are integers between 0 and 9. Then, the last two digits of U = NM are given by
u0 = n0m0 mod 10 and
u1 = n0m1 + n1m0 + floor(n0m0/10) mod 10
It is good practice to look into the circumstances under which m0 and m1 can be chosen so that u1 = u0. This implies:
m1n0 = n0m0  m0n1  floor(n0m0/10) mod 10
Now, if n0 is relatively prime to 10 (n0 = 1,3, or 7) then we have the fortuitous circumstance that, by varying m1, we can reach all possible values mod 10, so we are also completely free in how we pick m0..

Posted by FrankM
on 20150214 13:15:51 