The units digit of a perfect square is 6. What are the possible values of the tens digit?
The squre root has to be a 4 or 6 for the square to end in 6
4 16
6 36
14 196
16 256
24 576
26 676
34 1156
36 1296
44 1936
46 2116 So I'd say 1 3 9 5 or 7 for the tens' digit.....
squares that end in 6 must be of the form
\((10n\pm4)^2\) where n is an interger
\((10n\pm4)^2\\ =100n^2\pm80n+16\\ =100n^2+10\pm80n+6\\ =100n^2+10(1\pm8n)+6\\ \text{8n can end in any even number so }1\pm8n \text { can end in any odd number}\\ \text{So the 10s place value can have any odd digit.} \)