Let A=2^n+3^n+7^n
For a given positive integer n what is the last digit of A?

Last digits of 2^n have a cycle of four: 2, 4, 8, 6.

Last digits of 3^n have a cycle of four: 3, 9, 7, 1.

Last digits of 7^n have a cycle of four: 7, 9, 3, 1.

The total therefore has a cycle of four: 2, 2, 8, 8.

This means, depending on n mod 4 being 0 through 3, the values are 8, 2, 2 and 8 respectively.

Posted by Charlie on 2011-01-17 18:14:02