# Check if a number N can be expressed as the sum of powers of X or not

• Difficulty Level : Medium
• Last Updated : 02 Jun, 2021

Given two positive numbers N and X, the task is to check if the given number N can be expressed as the sum of distinct powers of X. If found to be true, then print “Yes”, Otherwise, print “No”.

Examples:

Input: N = 10, X = 3
Output: Yes
Explanation:
The given value of N(= 10) can be written as (1 + 9) = 30 + 32. Since all the power of X(= 3) are distinct. Therefore, print Yes.

Input: N= 12, X = 4
Output: No

Approach: The given problem can be solved by checking if the number N can be written in base X or not. Follow the steps below to solve the problem:

• Iterate a loop until the value of N is at least 0 and perform the following steps:
• Calculate the value of remainder rem when N is divided by X.
• If the value of rem is at least 2, then print “No” and return.
• Otherwise, update the value of N as N / X.
• After completing the above steps, if there doesn’t exist any termination, then print “Yes” as the result at N can be expressed in the distinct power of X.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; // Function to check if the number N// can be expressed as the sum of// different powers of X or notbool ToCheckPowerofX(int n, int x){    // While n is a positive number    while (n > 0) {         // Find the remainder        int rem = n % x;         // If rem is at least 2, then        // representation is impossible        if (rem >= 2) {            return false;        }         // Divide the value of N by x        n = n / x;    }     return true;} // Driver Codeint main(){    int N = 10, X = 3;    if (ToCheckPowerofX(N, X)) {        cout << "Yes";    }    else {        cout << "No";    }     return 0;}

## Java

 // Java program for the above approachimport java.io.*;import java.util.*; class GFG{ // Function to check if the number N// can be expressed as the sum of// different powers of X or notstatic boolean ToCheckPowerofX(int n, int x){         // While n is a positive number    while (n > 0)    {                 // Find the remainder        int rem = n % x;         // If rem is at least 2, then        // representation is impossible        if (rem >= 2)        {            return false;        }         // Divide the value of N by x        n = n / x;    }    return true;} // Driver Codepublic static void main (String[] args){    int N = 10, X = 3;    if (ToCheckPowerofX(N, X))    {        System.out.print("Yes");    }    else    {        System.out.print("No");    }}} // This code is contributed by sanjoy_62

## Python3

 # Python3 program for the above approach # Function to check if the number N# can be expressed as the sum of# different powers of X or notdef ToCheckPowerofX(n, x):         # While n is a positive number    while (n > 0):                 # Find the remainder        rem = n % x         # If rem is at least 2, then        # representation is impossible        if (rem >= 2):            return False         # Divide the value of N by x        n = n // x     return True # Driver Codeif __name__ == '__main__':         N = 10    X = 3         if (ToCheckPowerofX(N, X)):        print("Yes")    else:        print("No")         # This code is contributed by bgangwar59

## C#

 // C# program for the above approachusing System;class GFG{      // Function to check if the number N// can be expressed as the sum of// different powers of X or notstatic bool ToCheckPowerofX(int n, int x){         // While n is a positive number    while (n > 0)    {                 // Find the remainder        int rem = n % x;         // If rem is at least 2, then        // representation is impossible        if (rem >= 2)        {            return false;        }         // Divide the value of N by x        n = n / x;    }    return true;} // Driver codepublic static void Main(String []args){     int N = 10, X = 3;    if (ToCheckPowerofX(N, X))    {        Console.Write("Yes");    }    else    {        Console.Write("No");    }}}// This code is contributed by code_hunt,

## Javascript



Output:

Yes

Time Complexity: O(log N)
Auxiliary Space: O(1)

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