RMO 1993

This is how I did it:if you carefully notice the last two digits of the powers of 19 you will get the following pattern:19,49,79,09,39,69,99,29,59,89 and this will keep reapeating.From this we get that the last two digits of 19^93 are 79.Now,moving on to the last two digits of the powers of 13 to the power ofa no.which is a multiple of 3 you get the following pattern:97,17,37,57,77 and then it repeats.Which leads us to the conclusion that 13^99 will have 77 as its last two digits.Thus the last two digits of the no. in the question are 02.That implies that the number has to be 162.My God!!!!.

This problem can be solved through $mod$ function.

$19^{93}-13^{99} mod$ $299 = 148$ (Not divisible)

$19^{93}-13^{99} mod$ $173 = 94$ (Not divisible)

$19^{93}-13^{99} mod$ $204 = 198$ (Not divisible)

$19^{93}-13^{99} mod$ $162 = 0$ (Divisible)

Thus, the answer is $\boxed{162}$