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This is how I approach the problem
The given number is : aabb & the requirement is sqrt(aabb) = an integer.
aabb = aa00 + bb = aa*100 + bb = a*11*100 + b*11 = 11* (100a + b)
Because 11 is not a perfect square, then (100a + b) must have the factor 11 in them
11* (100a+b) = 11* 11 * ( 100a+b)/11
Now sqrt(aabb) will be equal to 11* sqrt((100a+b)/11)
In order for aabb to be a perfect square, (100a+b)/11 must also be a perfect square, and of course it must be an integer as well.
(100a + b)/11 = 9*a + (a+b)/11 is an interger , then a+b must be divisible to 11
a+ b =<18 ( as a&b =<9), so (a+ b)/11 = 1
Let's plug in all the possible a & b:
a=1 b =10, eliminate
a=2 b=9, then 9*a +(a+b)/11 = 2*9 + 1 = 19, not a perfect square, eliminate. Call 9*a + (a+b)/11 a name f(a,b) for short.
a=3 b=8, f(a,b) = 3*9 +1 = 28, eliminate for the same reason above.
a=4,b=7, f(a,b) = 37
a=5,b=6, f(a,b) = 46
a=6, f(a,b) = 55
a=7, f(a,b) = 64, that's it, 64 = 8^2, a perfect square.
a=8, f = 73
a=9, f= 82
In conclusion, there's only one 4-digit no. that matches the requirement.
It's when a =7, b = 4, the no. is 7744, and sqrt(7744) = 88
D is my answer.
