The difference between two positive integers is 56.
If their squares have both the same units digit and the same tens digit, find the sum of the two positive integers.

1. Find the minimal couple, complying with the above, and the corresponding sum.
2. Define formulas for additional solutions (both numbers and their sums).

broll solution is ok.
b-a=56 so b^2-a^2=(b-a)(b+a)=112a+3136 Last two digits of the sum are 00.
So 112a last two digits are 64, a=22+25k b=78+25k.
Edited on March 1, 2016, 1:46 am

Posted by armando on 2016-02-29 16:51:16