A Pre-RMO question! -13

Number Theory Level 1

How many two digit positive integer $N$ are there such that sum of $N$ and it's reverse is a perfect square?

Note:Reverse of $N$ means the number you get if you swipe the digits, for example if $N=21$ then reverse of $N=12$

The answer is 8.

3 solutions

Zakir Husain
Jun 12, 2020

Let $N=10p+q$ and set $A=$ { $x|x=y^2;y∈Z$ }

According to question, $(10p+q+10q+p)∈A$ $(11p+11q)∈A$ $(11(p+q))∈A$ $\Rightarrow p+q=11m;m∈A$ Now as maximum value of both $p$ and $q$ is $9$ $\therefore p+q<18$ Now as $p+q=11m;m∈A\Rightarrow m∈Z^+$ therefore $p+q$ is a multiple of $11$ . As $11$ and $0$ are the multiple of $11$ less than $18$ $\therefore p+q=11,p+q=0$ But as $0<p<10,0<q<10,p∈Z,q∈Z$ therefore $p+q≠0$ $\therefore p+q=11$ Now possible values of $p,q$ such that $0<p<10,0<q<10,p∈Z,q∈Z,p+q=11$ are

	$p$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$q$
$1$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$
$2$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\blue{92}$
$3$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\blue{83}$	$\red{\times}$
$4$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\blue{74}$	$\red{\times}$	$\red{\times}$
$5$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\blue{65}$	$\red{\times}$	$\red{\times}$	$\red{\times}$
$6$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\blue{56}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$
$7$		$\red{\times}$	$\red{\times}$	$\red{\times}$	$\blue{47}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$
$8$		$\red{\times}$	$\red{\times}$	$\blue{38}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$
$9$		$\red{\times}$	$\blue{29}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$	$\red{\times}$

Mahdi Raza
Jun 11, 2020

$N = 10a + b$
Reverse of $N = 10b + a$
Sum of $N$ and it's reverse $= 10 a + b + 10b + a = 11(a+b)$
For this to be a perfect square $a + b = 11k^2$ , but since $a,b$ are single digit numbers, their maximum is 18.
Thus we solve for $a+b = 11$ . There are $\boxed{8}$ such solutions

@Zakir Husain , more PRMO questions please

Mahdi Raza - 1 year ago

A Former Brilliant Member
Jun 11, 2020

Let such a number be $10a+b$ , where $0<a\leq 9$ and $0\leq b\leq 9$ are the digits of the number. Then

$11(a+b)$ must be a perfect square $\implies a+b$ must be a square multiple of $11$ . Since $a+b\leq 18$ , therefore $a+b=11$ . There are $\boxed 8$ such possible combinations of $(a, b)$ :

$(a, b) =(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)$ .

Can you make a note showing your solution, @Zakir Husain ?

A Former Brilliant Member - 1 year ago

@Yajat Shamji - Are you talking about solution to this problem?

Zakir Husain - 1 year ago

I've read @Mahdi Raza and @Alak Bhattacharya 's solutions, but I still don't understand it. I believe your solution will alleviate my doubts...

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – @Yajat Shamji , what did you not understand in particular?

Mahdi Raza - 1 year ago

@Mahdi Raza – Like, does the two-digit number $N$ 's sum has to be a perfect square when reversed as well? @Mahdi Raza , @Alak Bhattacharya , @Zakir Husain

A Former Brilliant Member - 1 year ago

@A Former Brilliant Member – Neither $N$ nor it's reverse has to be a perfect square. Instead, their sum has to be a perfect square.

Mahdi Raza - 1 year ago

@Mahdi Raza – Edited my comment - there was a mistake! @Mahdi Raza

A Former Brilliant Member - 1 year ago

A Pre-RMO question! -13

The answer is 8.

3 solutions

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