+239 The units digit of a perfect square is 6. What are the possible values of the tens digit?
+983 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.
