1. $$(BE)^2 = MPB$$, where B, E, M and P are distinct integers. Then M =
Write Comment
Comments
By: anil on 05 May 2019 02.31 am
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.
Terms And Service:We do not guarantee the accuracy of available data ..We Provide Information On Public Data.. Please consult an expert before using this data for commercial or personal use
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.