CAT 1998 | Question: 61

Question

$(BE)^{2} = MPB$, where $B, E, M$ & $P$ are distinct integers, then $M$?

• A. $2$
• B. $3$
• C. $9$
• D. None of these

Answer 1

There is only one possibility.

$19^{2}=361$

Here B=1

E=9

M=3

P=6

Here all numbers are distinct.

Hence,Option(B) 3 should be the correct choice.