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

  1. $2$
  2. $3$
  3. $9$
  4. None of these
asked in Quantitative Aptitude by (2.5k points)   | 16 views

1 Answer

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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.

answered by (228 points)  

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