Question 49

$$(BE)^2 = MPB$$, where B, E, M and P are distinct integers. Then M =

Solution

A digit number when squared produces a 3 digit number. This means that the number ranges from [10, 31].
First digit of $$BE^2$$ should be unit digit of $$E^2$$. But unit digit of $$E^2$$ is B. Look at the numbers and the unit digit of their square.

0-0, 1-1, 2-4, 3-9, 4-6, 5-5, 6-6, 7-9, 8-4, 9-1. Only 2-4, 3-9, 4-6, 7-9, 8-4 and 9-1 are kind of pairs we are looking after.But all the pairs except 9-1 produce a number greater than 31. Now, the number we can form from 9-1 is 19 whose square is 361 which satisfies all the condition we are looking for. This is the only such number.

Video Solution

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