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Algebra Level 3

let $x$ and $y$ be two 2-digit numbers such that $y$ is obtained by reversing the digits of $x$ . Suppose they also satisfy $x^2-y^2=m^2$ for some positive integer $m$ . Then the value of $x+y+m$ is:

The answer is 154.

1 solution

Adyasha Jena
Feb 24, 2016

let x be 10a+b....... so y= 10b+a.... x^{2}-y^{2}=m^{2} => (10a+b)^2-(10b+a)^2=m^2

on simplifying the above equation we get.......

99(a^2-b^2)=m^2

99(a^2-b^2) will be a square only when the value of a^2-b^2 is atleast 11

so a^2-b^2=11...... .(a+b)(a-b)=11

a+b=11.....and a-b=1

so we get a=6 and b=5

so x=65 and y=56 and hence m^2=1089

m=33

x+y+m=154

then the question should be asking for the minimum value of x+y+m.

Chirayu Bhardwaj - 5 years ago

Make it fast

The answer is 154.

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