# Minimum number of power terms with sum equal to n

Given two positive integer **n** and **x**. The task is to express n as sum of powers of x (x^{a1} + x^{a2} +…..+ x^{a3}) such that the number of powers of x (x^{a1}, x^{a2}, ….., x^{a3}) should be minimum. Print the minimum number of power of x used to make sum equal to n.

Examples:

Input : n = 5, x = 3 Output : 3 5 = 3^{0}+ 3^{0}+ 3^{1}. We use only 3 power terms of x { 3^{0}, 3^{0}, 3^{1}} Input : n = 13, x = 4 Output : 4 13 = 4^{0}+ 4^{1}+ 4^{1}+ 4^{1}. We use only four power terms of x. Input : n = 6, x = 1 Output : 6

If x = 1, then answer will be n only (n = 1 + 1 +…. n times).

The idea is to use Horner’s method. Any number n can be expressed as, n = x * a + b where 0 <= b <= x-1. Now since b is between 0 to x – 1, then b should be expressed as sum of x^{0} b times.

Further a can be decomposed in similar manner and so on.

Algorithm to solve this problem:

1. Initialize a variable ans to 0. 2. While n > 0 a) ans = ans + n % x b) n = n/x 3. Return ans.

Below is the implementation of above idea :

## C++

`// C++ program to calculate minimum number` `// of powers of x to make sum equal to n.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Return minimum power terms of x required` `int` `minPower(` `int` `n, ` `int` `x)` `{` ` ` `// if x is 1, return n since any power` ` ` `// of 1 is 1 only.` ` ` `if` `(x==1)` ` ` `return` `n;` ` ` `// Consider n = a * x + b where a = n/x` ` ` `// and b = n % x.` ` ` `int` `ans = 0;` ` ` `while` `(n > 0)` ` ` `{` ` ` `// Update count of powers for 1's added` ` ` `ans += (n%x);` ` ` `// Repeat the process for reduced n` ` ` `n /= x;` ` ` `}` ` ` `return` `ans;` `}` `// Driven Program` `int` `main()` `{` ` ` `int` `n = 5, x = 3;` ` ` `cout << minPower(n, x) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to calculate` `// minimum numberof powers of` `// x to make sum equal to n.` `class` `GFG` `{` ` ` `// Return minimum power` ` ` `// terms of x required` ` ` `static` `int` `minPower(` `int` `n, ` `int` `x)` ` ` `{` ` ` `// if x is 1, return n since` ` ` `// any power of 1 is 1 only.` ` ` `if` `(x==` `1` `)` ` ` `return` `n;` ` ` ` ` `// Consider n = a * x + b where` ` ` `// a = n/x and b = n % x.` ` ` `int` `ans = ` `0` `;` ` ` `while` `(n > ` `0` `)` ` ` `{` ` ` `// Update count of powers` ` ` `// for 1's added` ` ` `ans += (n % x);` ` ` ` ` `// Repeat the process for reduced n` ` ` `n /= x;` ` ` `}` ` ` ` ` `return` `ans;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `n = ` `5` `, x = ` `3` `;` ` ` `System.out.println(minPower(n, x));` ` ` `}` `}` `// This code is contributed by Anant Agarwal.` |

## Python3

`# Python program to` `# calculate minimum number` `# of powers of x to make` `# sum equal to n.` `# Return minimum power` `# terms of x required` `def` `minPower(n,x):` ` ` `# if x is 1, return` ` ` `# n since any power` ` ` `# of 1 is 1 only.` ` ` `if` `(x` `=` `=` `1` `):` ` ` `return` `n` ` ` ` ` `# Consider n = a * x + b where a = n/x` ` ` `# and b = n % x.` ` ` `ans ` `=` `0` ` ` `while` `(n > ` `0` `):` ` ` `# Update count of powers for 1's added` ` ` `ans ` `+` `=` `(n` `%` `x)` ` ` ` ` `# Repeat the process for reduced n` ` ` `n ` `/` `/` `=` `x` ` ` ` ` `return` `ans` ` ` `# Driver code` `n ` `=` `5` `x ` `=` `3` `print` `(minPower(n, x))` `# This code is contributed` `# by Anant Agarwal.` |

## C#

`// C# program to calculate` `// minimum numberof powers` `// of x to make sum equal` `// to n.` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Return minimum power` ` ` `// terms of x required` ` ` `static` `int` `minPower(` `int` `n, ` `int` `x)` ` ` `{` ` ` `// if x is 1, return n since` ` ` `// any power of 1 is 1 only.` ` ` `if` `(x == 1)` ` ` `return` `n;` ` ` ` ` `// Consider n = a * x + b where` ` ` `// a = n / x and b = n % x.` ` ` `int` `ans = 0;` ` ` `while` `(n > 0)` ` ` `{` ` ` `// Update count of` ` ` `// powers for 1's` ` ` `// added` ` ` `ans += (n % x);` ` ` ` ` `// Repeat the process` ` ` `// for reduced n` ` ` `n /= x;` ` ` `}` ` ` ` ` `return` `ans;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `n = 5, x = 3;` ` ` `Console.WriteLine(minPower(n, x));` ` ` `}` `}` `// This code is contributed by vt_m.` |

## PHP

`<?php` `// PHP program to calculate minimum number` `// of powers of x to make sum equal to n.` `// Return minimum power` `// terms of x required` `function` `minPower(` `$n` `, ` `$x` `)` `{` ` ` ` ` `// if x is 1, return n since` ` ` `// any power of 1 is 1 only.` ` ` `if` `(` `$x` `== 1)` ` ` `return` `$n` `;` ` ` `// Consider n = a * x + b` ` ` `// where a = n/x and b = n % x.` ` ` `$ans` `= 0;` ` ` `while` `(` `$n` `> 0)` ` ` `{` ` ` `// Update count of powers` ` ` `// for 1's added` ` ` `$ans` `+= (` `$n` `% ` `$x` `);` ` ` `// Repeat the process` ` ` `// for reduced n` ` ` `$n` `/= ` `$x` `;` ` ` `}` ` ` `return` `$ans` `;` `}` `// Driver Code` `$n` `= 5; ` `$x` `= 3;` `echo` `(minPower(` `$n` `, ` `$x` `));` `// This code is contributed by Ajit.` `?>` |

## Javascript

`<script>` `// JavaScript program to calculate minimum number` `// of powers of x to make sum equal to n.` `// Return minimum power terms of x required` `function` `minPower(n, x)` `{` ` ` `// if x is 1, return n since any power` ` ` `// of 1 is 1 only.` ` ` `if` `(x==1)` ` ` `return` `n;` ` ` `// Consider n = a * x + b where a = n/x` ` ` `// and b = n % x.` ` ` `let ans = 0;` ` ` `while` `(n > 0)` ` ` `{` ` ` `// Update count of powers for 1's added` ` ` `ans += (n%x);` ` ` `// Repeat the process for reduced n` ` ` `n = Math.floor(n / x);` ` ` `}` ` ` `return` `ans;` `}` `// Driven Program` ` ` `let n = 5, x = 3;` ` ` `document.write(minPower(n, x) + ` `"<br>"` `);` `// This code is contributed by Surbhi Tyagi.` `</script>` |

**Output: **

3

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