If (AB)2 = CCB then find minimum value of A+B+C. Here, AB is 2 digit and CCB is 3 digit number and A,B,C unequal, A,C not equals 0.

And for AB ^ 2 to have the last digit as B, we know that by inspection, B has to be 0, 1, 5 or 6.

(0 * 0, 1 * 1 = 1, 5 * 5 = 25, 6 * 6 = 36).

If we assume A = 1, B =5 , we get 15 * 15 = 225 which is of the format CCB.

Therefore A = 1, B = 5, C = 2.

If we assume A = 2, B = 1 we get 441 which is also of the format CCB.

Therefore A = 2, B = 1, C = 4 is also a solution.