Which of the following could be the last 2 digits of an integer that is a perfect square? What is your reasoning?

12
23
56
87
99

3 answers

1^2 and 9^2 end in 1
2^2 and 8^2 end in 4
3^2 and 7^2 end in 9
4^2 and 6^2 end in 6
5^2 ends in 5
So, perfect squares cannot end in 2,3,7,8
So the only possible choices are 56 and 99
Can you narrow it down any more than that?
I got that far, and I selected 99, but the answer is 56, and I'm not sure why. Could you please help me understand this?
well, 16^2 = 256