If A, B and C are distinct digits such that the square of the two

Responding to a pm:


Actually, the problem isn't very tough. Certainly more tedious than I would like but 2 mins is enough to solve it.

"the square of the two digit number AB equals the three digit number CCB"
The number is AB and its square is CCB - both end with the same digit.
This happens with only 4 digits: 0, 1, 5, 6
The square of a number ending with 0 will have two zeroes at the end and will be of the format C00 i.e. CBB (not CCB as given). Hence 0 is not possible.

Units digit 1 - 11, 21, 31 etc Larger numbers' square is 4 digit so ignore them.
Square of 11 is 121 - (Not Acceptable). Square of 21 = 441 (Acceptable). Square of 31 is 961 (Not acceptable)

Unit's digit 5 - 15, 25. Larger numbers' square is 4 digit (35^2 = 1225) so ignore them.
Square of 15 is 225 (Acceptable). Square of 25 is 625 (Not Acceptable).

We already have the sum as 441 + 225 = 666

Unit's digit 6 - 16, 26
Square of 16 is 256 (Not Acceptable). Square of 26 will be much larger such that when you add it to 666, the sum will exceed 1000 (not in options).

Hence, answer must be (B)