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# Counting numbers whose difference from reverse is a product of k in C++

We are given a range [l,r] and a number k. The goal is to find all the numbers between l and r (l<=number<=r) such that (reverse of that number)-(number) results into a product of k.

We will check this condition by starting from l to r, calculate the reverse of each number. Now subtract the number from its reverse and check if (absolute difference) %k==0. If yes then increment the count.

Let’s understand with examples.

**Input** − L=21, R=25, K=6

**Output** − Count of numbers − 2

**Explanation** −

The numbers their reverse and difference is: 21, 12, | 21-12 |=9, 9%6!=0 22, 22, | 22-22 |=0 0%6=0 count=1 23,32, | 32-23 | =9 9%6!=0 24,42, | 42-24 | =18 18%6=0 count=2 25,52, | 52-25 | =27 27%6!=0 Total numbers that meet the condition are 2 ( 22,24 )

**Input** − L=11, R=15, K=5

**Output** − Count of numbers − 1

**Explanation** −

The only number is 11 , | 11-11 | is 0 and 0%5=0

## Approach used in the below program is as follows

We take an integers L and R to define the range. And K for checking divisibilty.

Function countNumbers(int l, int r, int k) takes l,r and k as input and returns the count of numbers that meet the required condition.

Take the initial count as 0.

Take reverse of number rev=0.

Take remainder rem=0.

Starting from i= l to i= r.

Store current number i in num. And its rev=0.

Now reverse the number num, while(num>0). rem=num%10. rev=rev*10+rem. num=num/10.

After the end of while rev has reverse of i.

Calculate absolute difference of rev and original value i. If this difference | i-rev |%k==0. Then increment the count.

Do this for all numbers in the range.

Final value of count is returned as numbers whose difference with reverse is the product of k.

## Example

#include <iostream> using namespace std; int countNumbers(int l, int r, int k){ int rev = 0; int count=0; int rem=0; for (int i = l; i <= r; i++){ int num=i; rev=0; while (num > 0){ // reverse the number rem=num%10; rev = rev * 10 + rem; num /= 10; } if((abs(i-rev))%k==0) //original number is i and its reverse is rev { count++; } } return count; } int main(){ int L= 18, R = 24, K = 6; cout <<" Numbers whose difference with reverse is product of k:"<<countNumbers(L,R,K); return 0; }

## Output

If we run the above code it will generate the following output −

Numbers whose difference with reverse is product of k:4

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