Question 1182800
I will have to go with ikleyn's solution here.

The problem with greenestamps' solution is that he essentially assumes that a*b = 6^6, which is an incorrect assumption.

ikleyn's solution is correct, but the solution requires a bit of motivation combinatorially, especially when explaining 
(1+2+3+4+5+6+7)*(1+2+3+4+5+6+7) = 28^2 = 784.

In fact the problem can be generalized further to the number A^n*B^m and not only to 2^6*3^6.  
(So the bases 2 and 3 are irrelevant, and A and B can be any distinct positive integers both greater than 1.)

In this case the number of ordered pairs will be ((n+1)(n+2))/2*((m+1)(m+2))/2 = (n+1)(n+2)(m+1)(m+2)/4.