Repunit numbers

A number is a Repunit in base B if it can be represented as a string of three or more 1’s in a base >= 2.

Check if N is a Repunit number

Given an integer N, the task is to check if N is a Repunit number in base B.

Examples:

Input: N = 31, B = 5
Output: Yes
31 can be written as 111 base in 5

Input: N = 5, B = 2
Output: No
5 is 101 in base 2



Approach: We will count the number of one’s in the base B of a given number N and also count the number of digits in the base B of a given number N. If they are same, print “YES” else print “NO”.

For Example:

N = 31, B = 5
31 can be written as 111 base in 5, So number of one’s in base B of a given number N = 3 and number of digits in the base B of a given number N = 3
Since both are equal hence 31 is a Repunit number in base 5.

Below is the implementation of the above approach:

C++

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// C++ implementation  to check
// if a number is Repunit Number
  
#include <iostream>
#include <math.h>
using namespace std;
  
// Function to check if a number
// contains all the digits 0, 1, .., (b-1)
// an equal number of times
bool isRepunitNum(int n, int b)
{
    // to store number of digits of n
    // in base B
    int length = 0;
    // to count frequency of digit 1
    int countOne = 0;
    while (n != 0) {
        int r = n % b;
        length++;
        if (r == 1)
            countOne++;
        n = n / b;
    }
  
    // condition to check three or more 1's
    // and number of ones is equal to number
    // of digits of n in base B
    return countOne >= 3 && countOne == length;
}
  
// Driver Code
int main()
{
    // taking inputs
    int n = 31;
    int base = 2;
  
    // function to check
    if (isRepunitNum(n, base))
        cout << "Yes";
    else
        cout << "NO";
    return 0;
}

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Java

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// Java implementation to check
// if a number is Repunit Number
class GFG{
  
// Function to check if a number
// contains all the digits 0, 1, .., (b-1)
// an equal number of times
static boolean isRepunitNum(int n, int b)
{
    // to store number of digits of n
    // in base B
    int length = 0;
      
    // to count frequency of digit 1
    int countOne = 0;
    while (n != 0
    {
        int r = n % b;
        length++;
        if (r == 1)
            countOne++;
        n = n / b;
    }
  
    // condition to check three or more 1's
    // and number of ones is equal to number
    // of digits of n in base B
    return countOne >= 3 && 
           countOne == length;
}
  
// Driver Code
public static void main(String[] args) 
{
    // taking inputs
    int n = 31;
    int base = 2;
  
    // function to check
    if (isRepunitNum(n, base))
        System.out.print("Yes"); 
    else
        System.out.print("No");
}
}
  
// This code is contributed by rock_cool

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Python3

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# Python3 implementation to check
# if a number is Repunit Number
  
# Function to check if a number
# contains all the digits 0, 1, .., (b-1)
# an equal number of times
def isRepunitNum(n, b):
  
    # to store number of digits of n
    # in base B
    length = 0;
  
    # to count frequency of digit 1
    countOne = 0;
    while (n != 0):
        r = n % b;
        length += 1;
        if (r == 1):
            countOne += 1;
        n = n // b;
  
    # condition to check three or more 1's
    # and number of ones is equal to number
    # of digits of n in base B
    return countOne >= 3 and countOne == length;
  
# Driver Code
if __name__ == '__main__':
      
    # taking inputs
    n = 31;
    base = 2;
  
    # function to check
    if (isRepunitNum(n, base)):
        print("Yes");
    else:
        print("No");
  
# This code is contributed by 29AjayKumar

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C#

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// C# implementation to check
// if a number is Repunit Number
using System;
class GFG{
  
// Function to check if a number
// contains all the digits 0, 1, .., (b-1)
// an equal number of times
static bool isRepunitNum(int n, int b)
{
    // to store number of digits of n
    // in base B
    int length = 0;
      
    // to count frequency of digit 1
    int countOne = 0;
    while (n != 0) 
    {
        int r = n % b;
        length++;
        if (r == 1)
            countOne++;
        n = n / b;
    }
  
    // condition to check three or more 1's
    // and number of ones is equal to number
    // of digits of n in base B
    return countOne >= 3 && 
           countOne == length;
}
  
// Driver Code
public static void Main() 
{
    // taking inputs
    int n = 31;
    int base1 = 2;
  
    // function to check
    if (isRepunitNum(n, base1))
        Console.Write("Yes"); 
    else
        Console.Write("No");
}
}
  
// This code is contributed by Code_Mech

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Output:

Yes

Reference: http://www.numbersaplenty.com/set/repunit/

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