Given an integer **N**, the task is to find the smallest **N-digit** number which is a perfect fourth power.**Examples:**

Input:N = 2Output:16

Only valid numbers are 2^{4}= 16

and 3^{4}= 81 but 16 is the minimum.Input:N = 3Output:256

4^{4}= 256

**Approach:** It can be observed that for the values of **N = 1, 2, 3, …**, the series will go on like **1, 16, 256, 1296, 10000, 104976, 1048576, …** whose **N ^{th}** term will be

**pow(ceil( (pow(pow(10, (n – 1)), 1 / 4) ) ), 4)**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the smallest n-digit` `// number which is a perfect fourth power` `int` `cal(` `int` `n)` `{` ` ` `double` `res = ` `pow` `(` `ceil` `((` `pow` `(` `pow` `(10,` ` ` `(n - 1)), 1 / 4) )), 4);` ` ` `return` `(` `int` `)res;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 1;` ` ` `cout << (cal(n));` `}` `// This code is contributed by Mohit Kumar` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` `// Function to return the smallest n-digit` `// number which is a perfect fourth power` `static` `int` `cal(` `int` `n)` `{` ` ` `double` `res = Math.pow(Math.ceil((` ` ` `Math.pow(Math.pow(` `10` `,` ` ` `(n - ` `1` `)), ` `1` `/ ` `4` `) )), ` `4` `);` ` ` `return` `(` `int` `)res;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `1` `;` ` ` `System.out.println(cal(n));` `}` `}` `// This code is contributed by CodeMech` |

## Python3

`# Python3 implementation of the approach` `from` `math ` `import` `*` `# Function to return the smallest n-digit` `# number which is a perfect fourth power` `def` `cal(n):` ` ` `res ` `=` `pow` `(ceil( (` `pow` `(` `pow` `(` `10` `, (n ` `-` `1` `)), ` `1` `/` `4` `) ) ), ` `4` `)` ` ` `return` `int` `(res)` `# Driver code` `n ` `=` `1` `print` `(cal(n))` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` `// Function to return the smallest n-digit` `// number which is a perfect fourth power` `static` `int` `cal(` `int` `n)` `{` ` ` `double` `res = Math.Pow(Math.Ceiling((` ` ` `Math.Pow(Math.Pow(10,` ` ` `(n - 1)), 1 / 4) )), 4);` ` ` `return` `(` `int` `)res;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `n = 1;` ` ` `Console.Write(cal(n));` `}` `}` `// This code is contributed` `// by Akanksha_Rai` |

## Javascript

`<script>` `// Javascript implementation of the approach` `// Function to return the smallest n-digit` `// number which is a perfect fourth power` `function` `cal(n)` `{` ` ` `var` `res = Math.pow(Math.ceil((Math.pow(Math.pow(10,` ` ` `(n - 1)), 1 / 4) )), 4);` ` ` `return` `parseInt(res);` `}` `// Driver code` `var` `n = 1;` `document.write(cal(n));` `// This code is contributed by rutvik_56.` `</script>` |

**Output:**

1

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