The units digit of a perfect square is 6. What are the possible values of the tens digit?

Rollingblade
Apr 29, 2018

#1**+1 **

The second to last digit must be odd.

1, 3, 5, 7, and 9.

We can write the number in the form (10x + 6)^2 or (10x + 4)^2.

We then expand it.

100x^2 + 120x + 36 and 100x^2 + 80x + 16.

We don't need to 100x^2.

Then we factor into the form 10A + 6.

120x + 30 + 6 and 80x + 10 + 6.

Factoring, we have:

30(4x+1)+6 and 10(8x+1)+6.

The second to last digit of the perfect square is the last digit of 4x + 1 or 8x + 1

Since 8x is even, 8x + 1 must be odd.

Therefore the tens digit possible values or 1, 3, 5, 7, and 9.

Note: 0 is even, not odd.

I hope this helped,

Gavin.

GYanggg
Apr 29, 2018