CAT 1998 — QA Question 7

Since MPB is a three-digit number, and also the square of a two-digit number, it can have a maximum value of 961 viz. 31². This means that the number BE should be less than or equal to 31. ⇒ B can be 0, 1, 2 or 3. Since the last digit of MPB is also B, it can only be 0 or 1 (as none of the squares end in 2 or 3).

The only squares that end in 0 are 100, 400 and 900. But for this to occur the last digit of BE also has to be 0. Since E and B are distinct integers, both of them cannot be 0. Hence, B has to be 1. BE can be a number between 11 and 19 (as we have also ruled out 10), with its square also ending in 1.
Hence, the number BE can only be 11 or 19. 11² = 121. This is not possible as this will mean that M is also equal to 1. Hence, our actual numbers are 19² = 361.

Hence, M = 3.