# Find N numbers such that a number and its reverse are divisible by sum of its digits

Given a number N, the task is to print the first N numbers such that every number and the reverse of the number is divisible by its sum of digits.

Example:

Input: N = 4
Output: 1 2 3 4
Explanation:
The reverse of every single digit number is the same number. And, every number is divisible by itself.

Input: N = 12
Output: 1 2 3 4 5 6 7 8 9 10 12 18

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to iterate through every number from 1 and compute the sum of the digits. For every such number, check if the number and the reverse of the number are divisible by the sum or not. Therefore, the following steps are followed to compute the answer:

1. Initialize the counter to one and iterate through all the numbers one by one.
2. For every number, find the reverse of the number.
3. While finding the reverse of the number, compute the sum of the digits of the number.
4. Now, check if the number and the reverse of the number are divisible by the sum of its digits.
5. If it is, then increment the counter. Repeat the above steps until this counter is equal to N.

Below is the implementation of the above approach:

## C++

 `// C++ program to print the first N numbers ` `// such that every number and the reverse ` `// of the number is divisible by its ` `// sum of digits ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate the sum of digits ` `int` `digit_sum(``int` `n) ` `{ ` `    ``int` `sum = 0, m; ` ` `  `    ``// Loop to iterate through every ` `    ``// digit of the number ` `    ``while` `(n > 0) { ` `        ``m = n % 10; ` `        ``sum = sum + m; ` `        ``n = n / 10; ` `    ``} ` ` `  `    ``// Returning the sum of digits ` `    ``return` `(sum); ` `} ` ` `  `// Function to calculate the reverse ` `// of a number ` `int` `reverse(``int` `n) ` `{ ` `    ``int` `r = 0; ` ` `  `    ``// Loop to calculate the reverse ` `    ``// of the number ` `    ``while` `(n != 0) { ` `        ``r = r * 10; ` `        ``r = r + n % 10; ` `        ``n = n / 10; ` `    ``} ` ` `  `    ``// Return the reverse of the ` `    ``// number ` `    ``return` `(r); ` `} ` ` `  `// Function to print the first N numbers ` `// such that every number and the reverse ` `// of the number is divisible by its ` `// sum of digits ` `void` `operation(``int` `n) ` `{ ` `    ``int` `i = 1, a, count = 0, r; ` ` `  `    ``// Loop to continuously check and ` `    ``// generate number until there ` `    ``// are n outputs ` `    ``while` `(count < n) { ` ` `  `        ``// Variable to hold the sum of ` `        ``// the digit of the number ` `        ``a = digit_sum(i); ` ` `  `        ``// Computing the reverse of the ` `        ``// number ` `        ``r = reverse(i); ` ` `  `        ``// Checking if the condition satisfies. ` `        ``// Increment the count and print the ` `        ``// number if it satisfies. ` `        ``if` `(i % a == 0 && r % a == 0) { ` `            ``cout << i << ``" "``; ` `            ``count++; ` `            ``i++; ` `        ``} ` `        ``else` `            ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` ` `  `    ``operation(n); ` `} `

## Java

 `// Java program to print the first N numbers ` `// such that every number and the reverse ` `// of the number is divisible by its ` `// sum of digits ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Function to calculate the sum of digits ` `static` `int` `digit_sum(``int` `n) ` `{ ` `    ``int` `sum = ``0``, m; ` ` `  `    ``// Loop to iterate through  ` `    ``// every digit of the number ` `    ``while` `(n > ``0``) ` `    ``{ ` `        ``m = n % ``10``; ` `        ``sum = sum + m; ` `        ``n = n / ``10``; ` `    ``} ` ` `  `    ``// Returning the sum of digits ` `    ``return` `(sum); ` `} ` ` `  `// Function to calculate the  ` `// reverse of a number ` `static` `int` `reverse(``int` `n) ` `{ ` `    ``int` `r = ``0``; ` ` `  `    ``// Loop to calculate the  ` `    ``// reverse of the number ` `    ``while` `(n != ``0``)  ` `    ``{ ` `        ``r = r * ``10``; ` `        ``r = r + n % ``10``; ` `        ``n = n / ``10``; ` `    ``} ` ` `  `    ``// Return the reverse  ` `    ``// of the number ` `    ``return` `(r); ` `} ` ` `  `// Function to print the first N numbers ` `// such that every number and the reverse ` `// of the number is divisible by its ` `// sum of digits ` `static` `void` `operation(``int` `n) ` `{ ` `    ``int` `i = ``1``, a, count = ``0``, r; ` ` `  `    ``// Loop to continuously check and ` `    ``// generate number until there ` `    ``// are n outputs ` `    ``while` `(count < n)  ` `    ``{ ` ` `  `        ``// Variable to hold the sum  ` `        ``// of the digit of the number ` `        ``a = digit_sum(i); ` ` `  `        ``// Computing the reverse of the ` `        ``// number ` `        ``r = reverse(i); ` ` `  `        ``// Checking if the condition satisfies. ` `        ``// Increment the count and print the ` `        ``// number if it satisfies. ` `        ``if` `(i % a == ``0` `&& r % a == ``0``) ` `        ``{ ` `            ``System.out.print(i + ``" "``); ` `            ``count++; ` `            ``i++; ` `        ``} ` `        ``else` `            ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `n = ``10``; ` ` `  `    ``operation(n); ` `} ` `} ` ` `  `// This code is contributed by shivanisinghss2110 `

## Python3

 `# Pyhton3 program to print the first N numbers ` `# such that every number and the reverse ` `# of the number is divisible by its ` `# sum of digits ` ` `  `# Function to calculate the sum of digits ` `def` `digit_sum(n): ` `    ``sum` `=` `0` ` `  `    ``# Loop to iterate through every ` `    ``# digit of the number ` `    ``while` `(n > ``0``): ` `        ``m ``=` `n ``%` `10``; ` `        ``sum` `=` `sum` `+` `m; ` `        ``n ``=` `n ``/``/``10` ` `  `    ``# Returning the sum of digits ` `    ``return` `(``sum``) ` ` `  `# Function to calculate the reverse ` `# of a number ` `def` `reverse(n): ` `    ``r ``=` `0` ` `  `    ``# Loop to calculate the reverse ` `    ``# of the number ` `    ``while` `(n !``=` `0``): ` `        ``r ``=` `r ``*` `10` `        ``r ``=` `r ``+` `n ``%` `10` `        ``n ``=` `n ``/``/` `10` ` `  `    ``# Return the reverse of the ` `    ``# number ` `    ``return` `(r) ` ` `  `# Function to print the first N numbers ` `# such that every number and the reverse ` `# of the number is divisible by its ` `# sum of digits ` `def` `operation(n): ` `    ``i ``=` `1` `    ``count ``=` `0` ` `  `    ``# Loop to continuously check and ` `    ``# generate number until there ` `    ``# are n outputs ` `    ``while` `(count < n): ` `         `  `        ``# Variable to hold the sum of ` `        ``# the digit of the number ` `        ``a ``=` `digit_sum(i) ` ` `  `        ``# Computing the reverse of the ` `        ``# number ` `        ``r ``=` `reverse(i) ` ` `  `        ``# Checking if the condition satisfies. ` `        ``# Increment the count and print the ` `        ``# number if it satisfies. ` `        ``if` `(i ``%` `a ``=``=` `0` `and` `r ``%` `a ``=``=` `0``): ` ` `  `            ``print``(i, end ``=` `" "``) ` `            ``count ``+``=` `1` `            ``i ``+``=` `1` `        ``else``: ` `            ``i ``+``=` `1` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `     `  `    ``n ``=` `10` `    ``operation(n) ` ` `  `# This code is contributed by Samarth `

## C#

 `// C# program to print the first N numbers ` `// such that every number and the reverse ` `// of the number is divisible by its ` `// sum of digits ` `using` `System; ` `class` `GFG{ ` ` `  `// Function to calculate the sum of digits ` `static` `int` `digit_sum(``int` `n) ` `{ ` `    ``int` `sum = 0, m; ` ` `  `    ``// Loop to iterate through  ` `    ``// every digit of the number ` `    ``while` `(n > 0) ` `    ``{ ` `        ``m = n % 10; ` `        ``sum = sum + m; ` `        ``n = n / 10; ` `    ``} ` ` `  `    ``// Returning the sum of digits ` `    ``return` `(sum); ` `} ` ` `  `// Function to calculate the  ` `// reverse of a number ` `static` `int` `reverse(``int` `n) ` `{ ` `    ``int` `r = 0; ` ` `  `    ``// Loop to calculate the  ` `    ``// reverse of the number ` `    ``while` `(n != 0)  ` `    ``{ ` `        ``r = r * 10; ` `        ``r = r + n % 10; ` `        ``n = n / 10; ` `    ``} ` ` `  `    ``// Return the reverse  ` `    ``// of the number ` `    ``return` `(r); ` `} ` ` `  `// Function to print the first N numbers ` `// such that every number and the reverse ` `// of the number is divisible by its ` `// sum of digits ` `static` `void` `operation(``int` `n) ` `{ ` `    ``int` `i = 1, a, count = 0, r; ` ` `  `    ``// Loop to continuously check and ` `    ``// generate number until there ` `    ``// are n outputs ` `    ``while` `(count < n)  ` `    ``{ ` ` `  `        ``// Variable to hold the sum  ` `        ``// of the digit of the number ` `        ``a = digit_sum(i); ` ` `  `        ``// Computing the reverse of the ` `        ``// number ` `        ``r = reverse(i); ` ` `  `        ``// Checking if the condition satisfies. ` `        ``// Increment the count and print the ` `        ``// number if it satisfies. ` `        ``if` `(i % a == 0 && r % a == 0) ` `        ``{ ` `            ``Console.Write(i + ``" "``); ` `            ``count++; ` `            ``i++; ` `        ``} ` `        ``else` `            ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `n = 10; ` ` `  `    ``operation(n); ` `} ` `} ` ` `  `// This code is contributed by Code_Mech `

Output:

```1 2 3 4 5 6 7 8 9 10
```

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