A number theory problem by sathvik reddy

Number Theory Level 3

A 2 digit number 'n' has a property that the sum of n and the number obtained by reversing the digits is a 'perfect square'. Find the number of pairs of values which 'n' can have, i.e., the number of values of n divided by 2.

2 6 4 8

3 solutions

Pedro Cardoso
Aug 6, 2019

Nice first problem!

The number is of form $10a + b$ where $a$ and $b$ are the digits. This number gets its digits reversed, and gets added to itself, giving $(10a + b) + (10b + a) = 11(a+b)$ .

Since $11(a+b)$ has a factor of $11$ and it is a square obtained by adding $2$ -digit numbers, it must be equal to $11^2 = 121$ .

$11(a+b)=121 \Rightarrow a+b=11$

Since $a$ and $b$ are between $1$ and $9$ , the resulting pairs are: $2+9$ , $3+8$ , $4+7$ and $5+6$ , and the answer is $\boxed{4}$

Sathvik Reddy
Aug 6, 2019

let x, y be digits of n 10 x + y = n no. obtained by reversing digits is 10 y + x therefore, by adding them we get, 11(x+y) = perfect sq. since n is 2 digit no. max o f x+y =18 and min. val for 11(x+y)=121 x+y=11 (x,y)=(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2) so there are 8 values of n but they are asking no. of pairs i.e 4

Kyle T
Aug 6, 2019

<?php
for($i=10;$i<=99;$i++){
    $n = $i + strrev($i);
    if(floor(sqrt($n))==sqrt($n)){
        echo $i.'<br>';
    }
}
/*
pairs:
29 & 92
38 & 83
47 & 74
56 & 65
*/
?>

A number theory problem by sathvik reddy

3 solutions

1 pending report

Vote up reports you agree with