, 2003

Solution:

The answer is (B).

First of all we know that the number $AB$ and its square both have the same final digit $B$. Therefore $B$ must take one of the four values $0, 1, 5, 6$. We exclude the cases 0 and 1 since $C = B-1$ is a positive digit. In the other cases the relation $(AB)^2 = CAAB$ becomes:

1) $(A5)^2 = (10A+5)^2 = 100A^2 + 100A + 25 = 4AA5 = 4005 + 110A$ from which $10A^2 - A - 398 = 0$ which has no solution for $A = 1, 2, \ldots, 9$.

2) $(A6)^2 = (10A+6)^2 = 100A^2 + 120A + 36 = 5AA6 = 5006 + 110A$ from which $10A^2 + A - 497 = 0$ which has as integer solutions only $A = 7$.

Therefore the only solution for $AB$ is $AB = 76$.

SECOND SOLUTION

We must have $(10A+B)^2 = 100A^2 + 20AB + B^2 = 1000(B-1) + 100A + 10A + B$. Thus the final digit of the square of $B$ must be $B$, so $B = 0, 1, 5, 6$. Now $B$ must be greater than 1, because otherwise $B-1$ cannot be the initial digit of $(AB)^2$, so $B = 5, 6$. Now if $B = 5$ $A$ must be 6, because $(AB)^2$ must lie between 4000 and 5000; but $65^2 = 4225$ does not satisfy the required condition. If instead $B = 6$ $A$ must be 7, because $(AB)^2$ must lie between 5000 and 6000; and $76^2 = 5776$ satisfies the required condition.