# Find Nth smallest number that is divisible by 100 exactly K times

Given two numbers N and K . The task is to find N’th smallest number that is divided by 100 exactly K times.**Examples:**

Input :N = 12, K = 2Output :120000

120000 is divisible by 100 exactly 2 times and

is the 12 th smallest number also.Input :N = 1000, K = 2Output :10010000

**Approach:**

- First, find the smallest number that is divisible by 100 exactly K times. That is 2*K 0’s after 1 as 100 has two 0’s only.
- To find N’th smallest number, multiply N with the previous number we get after adding 2*k 0’s.
- Consider a case when N is divisible by 100 as if we multiply N with the previous number then the new number will have more than (2*k + 1) trailing 0’s that means it will divisible by 100 more than K times.
- Multiply that number with (N + 1). Use string as N and K can be very large that will not fit in integer limit.

Below is the implementation of above approach:

## C++

`// C++ implementation of above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the Nth smallest number` `string find_number(` `int` `N, ` `int` `K)` `{` ` ` `string r;` ` ` `// If N is divisible by 100 then we` ` ` `// multiply N + 1 otherwise, it will be` ` ` `// divisible by 100 more than K times` ` ` `if` `(N % 100 == 0) {` ` ` `N += 1;` ` ` `// convert integer to string` ` ` `r = to_string(N);` ` ` `}` ` ` `// if N is not divisible by 100` ` ` `else` `{` ` ` `// convert integer to string` ` ` `r = to_string(N);` ` ` `}` ` ` `// add 2*K 0's at the end to be divisible` ` ` `// by 100 exactly K times` ` ` `for` `(` `int` `i = 1; i <= K; i++)` ` ` `r += ` `"00"` `;` ` ` `return` `r;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 1000, K = 2;` ` ` `string ans = find_number(N, K);` ` ` `cout << ans << ` `"\n"` `;` ` ` `return` `0;` `}` |

## Java

`// Java implementation of above approach` `import` `java.util.*;` `class` `GFG` `{` `// Function to find the Nth smallest number` `static` `String find_number(` `int` `N, ` `int` `K)` `{` ` ` `String r;` ` ` `// If N is divisible by 100 then we` ` ` `// multiply N + 1 otherwise, it will be` ` ` `// divisible by 100 more than K times` ` ` `if` `(N % ` `100` `== ` `0` `)` ` ` `{` ` ` `N += ` `1` `;` ` ` `// convert integer to string` ` ` `r = String.valueOf(N);` ` ` `}` ` ` `// if N is not divisible by 100` ` ` `else` ` ` `{` ` ` `// convert integer to string` ` ` `r = String.valueOf(N);` ` ` `}` ` ` `// add 2*K 0's at the end to be divisible` ` ` `// by 100 exactly K times` ` ` `for` `(` `int` `i = ` `1` `; i <= K; i++)` ` ` `r += ` `"00"` `;` ` ` `return` `r;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `1000` `, K = ` `2` `;` ` ` `String ans = find_number(N, K);` ` ` `System.out.println(ans);` `}` `}` `/* This code is contributed by PrinciRaj1992 */` |

## Python3

`# Python3 implementation of above approach` `# Function to find the Nth smallest number` `def` `find_number(N, K):` ` ` ` ` `r ` `=` `""` ` ` `# If N is divisible by 100 then we` ` ` `# multiply N + 1 otherwise, it will be` ` ` `# divisible by 100 more than K times` ` ` `if` `(N ` `%` `100` `=` `=` `0` `):` ` ` `N ` `+` `=` `1` `;` ` ` `# convert integer to string` ` ` `r ` `=` `str` `(N)` ` ` `# if N is not divisible by 100` ` ` `else` `:` ` ` ` ` `# convert integer to string` ` ` `r ` `=` `str` `(N)` ` ` `# add 2*K 0's at the end to be divisible` ` ` `# by 100 exactly K times` ` ` `for` `i ` `in` `range` `(` `1` `, K ` `+` `1` `):` ` ` `r ` `+` `=` `"00"` ` ` `return` `r` `# Driver Code` `N ` `=` `1000` `K ` `=` `2` `;` `ans ` `=` `find_number(N, K)` `print` `(ans)` `# This code is contributed by Mohit Kumar` |

## C#

`// C# implementation of the approach` `using` `System;` ` ` `class` `GFG` `{` `// Function to find the Nth smallest number` `static` `String find_number(` `int` `N, ` `int` `K)` `{` ` ` `String r;` ` ` `// If N is divisible by 100 then we` ` ` `// multiply N + 1 otherwise, it will be` ` ` `// divisible by 100 more than K times` ` ` `if` `(N % 100 == 0)` ` ` `{` ` ` `N += 1;` ` ` `// convert integer to string` ` ` `r = N.ToString();` ` ` `}` ` ` `// if N is not divisible by 100` ` ` `else` ` ` `{` ` ` `// convert integer to string` ` ` `r = N.ToString();` ` ` `}` ` ` `// add 2*K 0's at the end to be divisible` ` ` `// by 100 exactly K times` ` ` `for` `(` `int` `i = 1; i <= K; i++)` ` ` `r += ` `"00"` `;` ` ` `return` `r;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 1000, K = 2;` ` ` `String ans = find_number(N, K);` ` ` `Console.WriteLine(ans);` `}` `}` `// This code is contributed by Rajput-Ji` |

## Javascript

`<script>` `// JavaScript implementation of above approach` `// Function to find the Nth smallest number` `function` `find_number(N, K)` `{` ` ` `var` `r;` ` ` `// If N is divisible by 100 then we` ` ` `// multiply N + 1 otherwise, it will be` ` ` `// divisible by 100 more than K times` ` ` `if` `(N % 100 == 0)` ` ` `{` ` ` `N += 1;` ` ` `// convert integer to string` ` ` `r = N.toString();` ` ` `}` ` ` `// if N is not divisible by 100` ` ` `else` ` ` `{` ` ` ` ` `// convert integer to string` ` ` `r = N.toString();` ` ` `}` ` ` `// add 2*K 0's at the end to be divisible` ` ` `// by 100 exactly K times` ` ` `for` `(` `var` `i = 1; i <= K; i++)` ` ` `r += ` `"00"` `;` ` ` `return` `r;` `}` ` ` `// Driver Code` `var` `N = 1000, K = 2;` `var` `ans = find_number(N, K);` `document.write(ans);` `// This code is contributed by Khushboogoyal499` `</script>` |

**Output:**

10010000

**Time Complexity: **O(K)

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