# Smallest number greater than or equal to N which is divisible by its non-zero digits

Given an integer **N**, the task is to find the smallest number greater than or equal to **N** such that it is divisible by all of its non-zero digits.

**Examples:**

Input:N = 31Output:33Explanation:33 is the smallest number satisfying the given condition.

At Unit’s place: 33%3 = 0

At One’s place: 33%3 = 0

Input:N = 30Output:30Explanation:30 is the smallest number satisfying the given condition.

At One’s place: 30%3 = 0

**Approach:** Smallest number which is divisible by all digits from 1 to 9 is equal to the LCM of (1, 2, 3, 4, 5, 6, 7, 8, 9) = 2520. Therefore, the multiples of **2520 **are also divisible by all digits from **1** to **9** implying that **(N + 2520)** will always satisfy the condition. Therefore, iterate in the range **[N, 2520 + N]** and check for the smallest number satisfying the given condition. Follow the steps below to solve the problem:

- Initialize
**ans**as**0**to store the smallest number greater than or equal to**N**such that it is divisible by all its non-zero digits. - Iterate over the range
**[N, N + 2520]**using the variable**i**.- Initialize a variable
**possible**as**1**to check if the current number**i**satisfies the given condition or not. - Get all non-zero digits of
**i**and check if i is divisible by each of them. If found to be true, then update**possible**to**1**, and update**ans**as**i**, and break out of the loop.

- Initialize a variable
- After the above steps, print the value of
**ans**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the smallest number` `// greater than or equal to N such that` `// it is divisible by its non-zero digits` `void` `findSmallestNumber(` `int` `n)` `{` ` ` `// Iterate in range[N, N + 2520]` ` ` `for` `(` `int` `i = n; i <= (n + 2520); ++i) {` ` ` `// To check if the current number` ` ` `// satisfies the given condition` ` ` `bool` `possible = 1;` ` ` `// Store the number in a temporary` ` ` `// variable` ` ` `int` `temp = i;` ` ` `// Loop until temp > 0` ` ` `while` `(temp) {` ` ` `// Check only for non zero digits` ` ` `if` `(temp % 10 != 0) {` ` ` `// Extract the current digit` ` ` `int` `digit = temp % 10;` ` ` `// If number is divisible` ` ` `// by current digit or not` ` ` `if` `(i % digit != 0) {` ` ` `// Otherwise, set` ` ` `// possible to 0` ` ` `possible = 0;` ` ` `// Break out of the loop` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Divide by 10` ` ` `temp /= 10;` ` ` `}` ` ` `if` `(possible == 1) {` ` ` `cout << i;` ` ` `return` `;` ` ` `}` ` ` `}` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 31;` ` ` `// Function Call` ` ` `findSmallestNumber(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` ` ` `class` `GFG{` ` ` `// Function to find the smallest number` `// greater than or equal to N such that` `// it is divisible by its non-zero digits` `static` `void` `findSmallestNumber(` `int` `n)` `{` ` ` ` ` `// Iterate in range[N, N + 2520]` ` ` `for` `(` `int` `i = n; i <= (n + ` `2520` `); ++i)` ` ` `{` ` ` ` ` `// To check if the current number` ` ` `// satisfies the given condition` ` ` `int` `possible = ` `1` `;` ` ` ` ` `// Store the number in a temporary` ` ` `// variable` ` ` `int` `temp = i;` ` ` ` ` `// Loop until temp > 0` ` ` `while` `(temp != ` `0` `)` ` ` `{` ` ` ` ` `// Check only for non zero digits` ` ` `if` `(temp % ` `10` `!= ` `0` `)` ` ` `{` ` ` ` ` `// Extract the current digit` ` ` `int` `digit = temp % ` `10` `;` ` ` ` ` `// If number is divisible` ` ` `// by current digit or not` ` ` `if` `(i % digit != ` `0` `)` ` ` `{` ` ` ` ` `// Otherwise, set` ` ` `// possible to 0` ` ` `possible = ` `0` `;` ` ` ` ` `// Break out of the loop` ` ` `break` `;` ` ` `}` ` ` `}` ` ` ` ` `// Divide by 10` ` ` `temp /= ` `10` `;` ` ` `}` ` ` ` ` `if` `(possible == ` `1` `)` ` ` `{` ` ` `System.out.println(i);` ` ` `return` `;` ` ` `}` ` ` `}` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `31` `;` ` ` ` ` `// Function Call` ` ` `findSmallestNumber(N);` `}` `}` `// This code is contributed by susmitakundugoaldanga` |

## Python3

`# Python3 program for the above approach` `# Function to find the smallest number` `# greater than or equal to N such that` `# it is divisible by its non-zero digits` `def` `findSmallestNumber(n):` ` ` `# Iterate in range[N, N + 2520]` ` ` `for` `i ` `in` `range` `(n, n ` `+` `2521` `):` ` ` `# To check if the current number` ` ` `# satisfies the given condition` ` ` `possible ` `=` `1` ` ` `# Store the number in a temporary` ` ` `# variable` ` ` `temp ` `=` `i` ` ` `# Loop until temp > 0` ` ` `while` `(temp):` ` ` `# Check only for non zero digits` ` ` `if` `(temp ` `%` `10` `!` `=` `0` `):` ` ` `# Extract the current digit` ` ` `digit ` `=` `temp ` `%` `10` ` ` `# If number is divisible` ` ` `# by current digit or not` ` ` `if` `(i ` `%` `digit !` `=` `0` `):` ` ` `# Otherwise, set` ` ` `# possible to 0` ` ` `possible ` `=` `0` ` ` `# Break out of the loop` ` ` `break` ` ` `# Divide by 10` ` ` `temp ` `/` `/` `=` `10` ` ` `if` `(possible ` `=` `=` `1` `):` ` ` `print` `(i, end ` `=` `"")` ` ` `return` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `31` ` ` `# Function Call` ` ` `findSmallestNumber(N)` ` ` `# This code is contributed by AnkThon` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the smallest number` `// greater than or equal to N such that` `// it is divisible by its non-zero digits` `static` `void` `findSmallestNumber(` `int` `n)` `{` ` ` ` ` `// Iterate in range[N, N + 2520]` ` ` `for` `(` `int` `i = n; i <= (n + 2520); ++i)` ` ` `{` ` ` ` ` `// To check if the current number` ` ` `// satisfies the given condition` ` ` `int` `possible = 1;` ` ` `// Store the number in a temporary` ` ` `// variable` ` ` `int` `temp = i;` ` ` `// Loop until temp > 0` ` ` `while` `(temp != 0)` ` ` `{` ` ` ` ` `// Check only for non zero digits` ` ` `if` `(temp % 10 != 0)` ` ` `{` ` ` ` ` `// Extract the current digit` ` ` `int` `digit = temp % 10;` ` ` `// If number is divisible` ` ` `// by current digit or not` ` ` `if` `(i % digit != 0)` ` ` `{` ` ` ` ` `// Otherwise, set` ` ` `// possible to 0` ` ` `possible = 0;` ` ` `// Break out of the loop` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `// Divide by 10` ` ` `temp /= 10;` ` ` `}` ` ` `if` `(possible == 1)` ` ` `{` ` ` `Console.Write(i);` ` ` `return` `;` ` ` `}` ` ` `}` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 31;` ` ` `// Function Call` ` ` `findSmallestNumber(N);` `}` `}` `// This code is contributed by shivanisinghss2110` |

## Javascript

`<script>` `// javascript program for the above approach` `// Function to find the smallest number` `// greater than or equal to N such that` `// it is divisible by its non-zero digits` `function` `findSmallestNumber(n)` `{` ` ` ` ` `// Iterate in range[N, N + 2520]` ` ` `for` `(i = n; i <= (n + 2520); ++i)` ` ` `{` ` ` ` ` `// To check if the current number` ` ` `// satisfies the given condition` ` ` `var` `possible = 1;` ` ` ` ` `// Store the number in a temporary` ` ` `// variable` ` ` `var` `temp = i;` ` ` ` ` `// Loop until temp > 0` ` ` `while` `(temp != 0)` ` ` `{` ` ` ` ` `// Check only for non zero digits` ` ` `if` `(temp % 10 != 0)` ` ` `{` ` ` ` ` `// Extract the current digit` ` ` `var` `digit = temp % 10;` ` ` ` ` `// If number is divisible` ` ` `// by current digit or not` ` ` `if` `(i % digit != 0)` ` ` `{` ` ` ` ` `// Otherwise, set` ` ` `// possible to 0` ` ` `possible = 0;` ` ` ` ` `// Break out of the loop` ` ` `break` `;` ` ` `}` ` ` `}` ` ` ` ` `// Divide by 10` ` ` `temp = parseInt(temp / 10);` ` ` `}` ` ` ` ` `if` `(possible == 1)` ` ` `{` ` ` `document.write(i);` ` ` `return` `;` ` ` `}` ` ` `}` `}` ` ` `// Driver code` `var` `N = 31;` `// Function Call` `findSmallestNumber(N);` `// This code is contributed by shikhasingrajput` `</script>` |

**Output:**

33

**Time Complexity:** O(2520*log_{10}N)**Auxiliary Space:** O(1)

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