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

• [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
• 11 years agoHelpfull: Yes(2) No(0)
• 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.
• 11 years agoHelpfull: Yes(1) No(0)
• (BE)^2 is a three digit no. So 10
• 11 years agoHelpfull: Yes(1) No(0)