The last two digits of a perfect square N, having at least 3 digits, is 09 (in this order).

Is the hundreds digit of N always even? Give reasons for your answer.

x²= N
For N to end in 9, x must end in 3 or 7.

For N to end in 09, x must be of the form 50n±3
(50n±3)²=2500n²±300n+9
Notice the hundreds place digits are both odd.  Since n² and n also have the same parity, the sum will always be even.

If you check
(50n±13)² ends in 69
(50n±23)² ends in 29
(50n±33)² ends in 89
(50n±43)² ends in 49

Posted by Jer on 2015-08-18 08:51:44