Problem Solving - Very Challenging?

Question

If the tens digit of a perfect square is 7, how many possible values can the units digit have?

Answer 1

the numer

= 10 x + y

now y < 10

(10x+y)^2 = 100x^2 + 20 x + y^2

for the 10s digit to be 7 y^2 has to have an odd digit in 10s position

taje the values 0 ( 0) 1(01),2(04)3(09),4(16), 5( 25) , 6(36), 7(49),8(64(, 9(81)

so 4 and 6 are choices of square root and 6 of the square

now we need to prove if some thing does not satisy.

but we have 24^2 = 576 and 26^2 = 676

so both satisfy

so choices are 2 digits for the square root   4 and 6 and choice of number is 6

so only solution 6

Answer 2

Just 6 (one possible value)

For example 24^2=576

Answer 3

0