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Smallest multiple of N with exactly N digits in its Binary number representation
  • Last Updated : 07 Apr, 2021
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Given a positive integer N, the task is to find the smallest multiple of N with exactly N digits in its binary number representation.
Example: 
 

Input: N = 3 
Output:
Explanation: 
6 is the smallest multiple of 3 and has length also 3(110) in binary.
Input: N = 5 
Output: 20 
Explanation: 
6 is the smallest multiple of 5 and has length also 5(10100) in binary. 
 

 

Approach: The idea is to make an observation. 
 

  • If we observe carefully a series will be formed as 1, 2, 6, 8, 20, …
  • The N-th term in the series would be:
     

N*\lceil 2^\frac{N-1}{N} \rceil
 



  • Therefore, the number N is taken as the input and the above formula is implemented.

Below is the implementation of the above approach:
 

C++




// C++ program to find smallest
// multiple of n with exactly N
// digits in Binary number System.
 
#include <iostream>
#include <math.h>
using namespace std;
 
// Function to find smallest multiple
// of n with exactly n digits
// in Binary number representation.
void smallestNumber(int N)
{
    cout << N * ceil(pow(2,
                         (N - 1))
                     / N);
}
 
// Driver code
int main()
{
    int N = 3;
    smallestNumber(N);
 
    return 0;
}

Java




// Java program to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
class GFG{
 
// Function to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
static void smallestNumber(int N)
{
    System.out.print(N * Math.ceil
                        (Math.pow(2, (N - 1)) / N));
}
 
// Driver code
public static void main(String[] args)
{
    int N = 3;
     
    smallestNumber(N);
}
}
 
// This code is contributed by shubham

Python3




# Python3 program to find smallest
# multiple of n with exactly N
# digits in Binary number System.
from math import ceil
 
# Function to find smallest multiple
# of n with exactly n digits
# in Binary number representation.
def smallestNumber(N):
    print(N * ceil(pow(2, (N - 1)) / N))
 
# Driver code
N = 3
smallestNumber(N)
 
# This code is contributed by Mohit Kumar

C#




// C# program to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
using System;
 
class GFG{
 
// Function to find smallest
// multiple of n with exactly N
// digits in Binary Number System.
static void smallestNumber(int N)
{
    Console.Write(N * Math.Ceiling(
                      Math.Pow(2, (N - 1)) / N));
}
     
// Driver code
public static void Main(string[] args)
{
    int N = 3;
         
    smallestNumber(N);
}
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
// Javascript program to find smallest
// multiple of n with exactly N
// digits in Binary number System.
 
// Function to find smallest multiple
// of n with exactly n digits
// in Binary number representation.
function smallestNumber(N)
{
    document.write(N * parseInt(Math.ceil(Math.pow(2,
                                 (N - 1))
                             / N)));
}
 
// Driver code
let N = 3;
smallestNumber(N);
 
// This code is contributed by rishavmahato348.
</script>
Output: 
6

 

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