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

 Apr 29, 2018
 #1
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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.

 Apr 29, 2018
 #2
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Thanks

 Apr 29, 2018
edited by Guest  Apr 29, 2018
 #3
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Your welcome!

 

Did you have the same question too?

GYanggg  Apr 29, 2018

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