Latest self Aptitude Question SOLUTION: (BE)^2=MPB, WHERE M,P,B,E ARE THE DISTINCT INTEGERS, THN M= ????

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(#M40025717) SELF question math Keep an EYE

(BE)^2=MPB, WHERE M,P,B,E ARE THE DISTINCT INTEGERS, THN M= ????

VIKASH KUMAR SHARMA (10 years ago)

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Amit kumar  (10 years ago)

[Amit kumar]
(BE)^2 is a three digit no. So 10 < BE < 31 => B must be 1,2 or 3
(BE)^2=MPB the last digit of perfect square should not be 2,3,7and 8.
So B = 1
so E must be 1 or 9, but all numbers are distinct so E=9
(BE)^2=MPB
(19)^2=361
B=1
E=9
M=3
P=6

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kumar sundram  (10 years ago)

here 1st(10th place) digit of two digit no. is equal to the last(unit place) digit of the square of the two digit no., which is a three digit no.
since, 2,3,5,7 & 8 is not the unit place of the square of any no. so, B is not equal to 2,3,5,7 & 8.
Now, assume, B=1, then E=1 or 9.since all digit are distinct.Therefore, E=9.
again, if B=1 & E=9 then M=3[(19)^2=361].
again, assume B=4 then E=2 or 8.
now, when B=4 & E=2 then (42)^2=1764 which is a four digit no. this is not follow the condition.
Thus, M=3.

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Amit kumar  (10 years ago)

(BE)^2 is a three digit no. So 10

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