Smallest N digit number which is a perfect fourth power
Given an integer N, the task is to find the smallest N-digit number which is a perfect fourth power.
Examples:
Input: N = 2
Output: 16
Only valid numbers are 24 = 16
and 34 = 81 but 16 is the minimum.
Input: N = 3
Output: 256
44 = 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 Nth 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 |
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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 |
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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 = 1print(cal(n)) |
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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 |
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Output:
1
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