manpreetsingh86 wrote:

Let p and q be two digit integers such that q is obtained by reversing the digits of p. The integers p and q satisfy the equation \(p^2 - q^2 = r^2\) for some positive integer r. what is the value of p+q+r ?

A) 88

B) 112

C) 116

D) 144

E) 154

let p = 10A+B

Then q = 10B+A

given that

\(p^2 - q^2 = r^2\) for positive integer r.

plugging in values of p and q we get

\((10A+B)^2-(10B+A)^2 = r^2\)

or \(100A^2 + B^2 + 20AB -100B^2 - A^2 - 20AB = r^2\)

or \(99A^2-99B^2=r^2\)

or \(9*11*(A+B)(A-B)=r^2\) ......(1)

For r to be integer, \(\sqrt{r^2}\) should be integer.

So, LHS of (1) should be perfect square.

9 is already a square. Either A+B or A-B should be equal to 11 to get square of 11.

A-B cannot be 11 as difference between 2 single digit numbers cannot be a two digit number.

So, A+B = 11 and A,B can be (6,5),(7,4),(8,3),(9,2)

Also A-B should be a perfect square. A-B can be equal to 1 or 4. If you see the values of A and B from above sets,A-B cannot be 4 in any case. So, A-B should be 1.

This is possible only when A = 6 and B = 5

plugging in these values of A and B in (1), we get r=33

Also, p=65 and q=56

So,

p+q+r= 65+56+33 = 154

Answer:-E