Minimum integer such that it leaves a remainder 1 on dividing with any element from the range [2, N]

Given an integer N, the task is to find the minimum possible integer X such that X % M = 1 for all M from the range [2, N]

Examples:

Input: N = 5
Output: 61
61 % 2 = 1
61 % 3 = 1
61 % 4 = 1
61 % 5 = 1

Input: N = 2
Output: 3



Approach: Find the lcm of all the integers from the range [2, N] and store it in a variable lcm. Now we know that lcm is the smallest number which is divisible by all the elements from the range [2, N] and to make it leave a remainder of 1 on every division, just add 1 to it i.e. lcm + 1 is the required answer.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the smallest number
// which on dividing with any element from
// the range [2, N] leaves a remainder of 1
long getMinNum(int N)
{
    // Find the LCM of the elements
    // from the range [2, N]
    int lcm = 1;
    for (int i = 2; i <= N; i++)
        lcm = ((i * lcm) / (__gcd(i, lcm)));
  
    // Return the required number
    return (lcm + 1);
}
  
// Driver code
int main()
{
    int N = 5;
  
    cout << getMinNum(N);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG {
  
    // Function to return the smallest number
    // which on dividing with any element from
    // the range [2, N] leaves a remainder of 1
    static long getMinNum(int N)
    {
        // Find the LCM of the elements
        // from the range [2, N]
        int lcm = 1;
        for (int i = 2; i <= N; i++)
            lcm = ((i * lcm) / (__gcd(i, lcm)));
  
        // Return the required number
        return (lcm + 1);
    }
  
    static int __gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }
  
    // Driver code
    public static void main(String args[])
    {
        int N = 5;
        System.out.println(getMinNum(N));
    }
}
  
// This code has been contributed by 29AjayKumar

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
from math import gcd
  
# Function to return the smallest number 
# which on dividing with any element from 
# the range [2, N] leaves a remainder of 1 
def getMinNum(N) : 
  
    # Find the LCM of the elements 
    # from the range [2, N] 
    lcm = 1
    for i in range(2, N + 1) :
        lcm = ((i * lcm) // (gcd(i, lcm))); 
  
    # Return the required number 
    return (lcm + 1); 
  
  
# Driver code 
if __name__ == "__main__"
  
    N = 5
  
    print(getMinNum(N)); 
  
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG {
  
    // Function to return the smallest number
    // which on dividing with any element from
    // the range [2, N] leaves a remainder of 1
    static long getMinNum(int N)
    {
        // Find the LCM of the elements
        // from the range [2, N]
        int lcm = 1;
        for (int i = 2; i <= N; i++)
            lcm = ((i * lcm) / (__gcd(i, lcm)));
  
        // Return the required number
        return (lcm + 1);
    }
  
    static int __gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return __gcd(b, a % b);
    }
  
    // Driver code
    public static void Main()
    {
        int N = 5;
        Console.WriteLine(getMinNum(N));
    }
}
  
// This code has been contributed by anuj_67..

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach
  
// Function to return the smallest number
// which on dividing with any element from
// the range [2, N] leaves a remainder of 1
function getMinNum($N)
{
    // Find the LCM of the elements
    // from the range [2, N]
    $lcm = 1;
    for ($i = 2; $i <= $N; $i++)
        $lcm = (($i * $lcm) / (__gcd($i, $lcm)));
  
    // Return the required number
    return ($lcm + 1);
}
  
function __gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return __gcd($b, $a % $b);
}
  
// Driver code
  
$N = 5;
echo (getMinNum($N));
      
// This code has been contributed by ajit....
?>

chevron_right


Output:

61


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, 29AjayKumar, vt_m, jit_t



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.