# how many two-digit positive integers n have the property that the sum of n and the number obtained by reversing the order of the digits of n is a perfect square?

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## How many two digit positive integers N have the property that the sum of N and number obtained by reversing the order of the digits of N is a perfect square.

How many two digit positive integers N have the property that the sum of N and number obtained by reversing the order of the digits of N is a perfect square.

Byju's Answer Standard XII Mathematics

Squaring an Inequality

How many two ... Question

How many two digit positive integers N have the property that the sum of N and number obtained by reversing the order of the digits of N is a perfect square.

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Let the two digits be x and y.

N=10x+y;

Let M be the number obtained on reversing the digits.

M=10y+x; N+M=10x+y+10y+x= =11x+11y=11 (x+y);

N+M is a perfect square; i.e. 11 (x+y) is a perfect square; i.e. 11 divides the perfect square

only 121 is divisible by 11.

Hence M+N=121 i.e. 11 (x+y)=121

x+y=11 and x, y are singel digit integers.

The integer solutions to the above equation are (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2).

Hence there are eight such two digits numbers such that the sum of the number and its reverse is a perfect square. They are 29, 38, 47, 56, 65, 74, 83, 92.

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Squaring an Inequality

Standard XII Mathematics

## How many two

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## How many two- digits positive integers N have the property that the sum of N and the number obtained by reversing the order of the digits of is a perfect square

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Updated on : 2022-09-05

Let units digit x and tens digit be ySolution Verified by Toppr Number=10y+x

Reverse number=10x+y

Sum =11x+11y =11(x+y)

For sum to be square

x+y=11 ∴ Number of ways= 7+2−1 C 2−1 = 8 C 1 =8

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## How many 2

Answer (1 of 4): Firstly, I'd like to clear whether I've understood the question properly. We have to find how many 2 digit numbers which on adding with their mirror images (obtained by reversing the order of the digits) give perfect squares, are there. Positive integers are just whole numbers g...

How many 2-digit positive integers have the property that the sum of N and the number obtained reversing the order of the digits is a perfect square?

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Let the two digits be x and y.

N=10x+y;

Let M be the number obtained on reversing the digits.

M=10y+x;

N+M=10x+y+10y+x=11x+11y=11 (x+y);

N+M is a perfect square; i.e. 11 (x+y) is a perfect square; i.e. 11 divides the perfect square

The maximum value of N+M can be 99+99=198;

Consider the perfect squares series 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196. The next perfect square is 225. We need not go beyond 196 as the maximum value of N+M is 198.

In the above aeries of perfect squares, only 121 is divisible by 11.

Hence M+N=121 i.e. 11 (x+y)=121 i.e. x+y=11 and x, y are singe digit integers

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Palash Saha 5y

Firstly, I'd like to clear whether I've understood the question properly.

We have to find how many 2 digit numbers which on adding with their mirror images (obtained by reversing the order of the digits) give perfect squares, are there. Positive integers are just whole numbers greater than zero. So we are hunting for the number (quantity) of the numbers, between 9 and 100, which on adding with their mirror images give perfect squares.

Right? Let me continue (please inform me if I'm wrong).

We can express any double digit number by 10x+y where x and y are 1, 2,….,8, 9, 0 (x can't be 0 as the numbe

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Anmol

BS-MS in Fundamental Physics, Indian Institute of Science Education and Research, Pune (IISER-P) (Graduated 2020)Author has 82 answers and 407.2K answer viewsUpdated 5y

Let Sum of the number and it's reverse: (10x+y)+(10y+x) = 11(x+y)

For positive interger z the Given Condition:

11(x+y)= z 2 11(x+y)=z2 =1 =1 1(x+y)= 11 2 1(x+y)=112 = x+y=11

So All solutions for (x,y) are:

(2,9), (3,8), (4,7), (5,6)

All such two digit numbers are:

29 and 92 38 and 83 47 and 74 56 and 65

Edit: To the People asking Why

z=11 z=11 ?

Well, The equation is:

11(x+y)= z 2 . 11(x+y)=z2.

Stare at the equation for a moment!

Your aim is to make 11(x+y) 11(x+y) a full square. 11(x+y) 11(x+y)

is a product of two numbers:

11 11 and (x+y). (x+y).

You see? You already have one number 11. If other unknown number (x+y) is 11 too, then the product will become full square. Simple!

P Nikhil Kumar

B.Tech (ECE) from National Institute of Technology, Patna (Graduated 2022)Author has 219 answers and 629.4K answer views4y

There are 8 such numbers.

29, 38, 47... Gowtham Naidu

Software Engineer (2019–present)Author has 56 answers and 712.4K answer views3y

Related

What is the missing Fibonacci number 144 __ 377 610?

The Fibonacci series is the type of series in which the number is addition of last two numbers.

In Fibonacci series the first numbers can be anything random but the 3rd number will be the sum of 1st and 2nd number and 4th number will sum of 2nd and 3rd number .

So,in this fibonacci series

144 ,___ ,377, 610.

The 3rd number is 377 which will be the sum of 1st number 144 and 2nd number .

So,

144 + 2nd number =377

2nd number =233

So, the sequence will become like 144,233,377,610.

(For recheck we can add 2nd number and 3rd number to get 4th number ie. 233+377 which gives 610 as the answer so the missing nu

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Bernard Montaron

Studied Mathematics & Counting (Combinatorics) at Pierre and Marie Curie University (Graduated 1980)Author has 1.9K answers and 601.4K answer viewsUpdated 1y

Related

What is the sum of the digits of the positive integer n such that the number 1! · 2! · 3! · · · 2019! · 2020! /n! Is a perfect square?

This is a great question! Thank you for asking it on Quora.

The answer is 2, and the number is

n=1010 n=1010

. Here is how it goes:

Guys, does anyone know the answer?