Nearest smaller number to N having multiplicative inverse under modulo N equal to that number

Given a prime number N, the task is to find the closest smaller number than N such that modulo multiplicative inverse of a number under modulo N is equal to the number itself.

Examples:

Input: N = 7
Output: 6
Explanation:
Modulo multiplicative inverse of all possible natural numbers from 1 to less than N are:
Modulo multiplicative inverse of 1 under modulo N(=7) is 1.
Modulo multiplicative inverse of 2 under modulo N(=7) is 4.
Modulo multiplicative inverse of 3 under modulo N(=7) is 5.
Modulo multiplicative inverse of 4 under modulo N(=7) is 2.
Modulo multiplicative inverse of 5 under modulo N(=7) is 3.
Modulo multiplicative inverse of 6 under modulo N(=7) is 6.
Therefore, the nearest smaller number to N(= 7) having modulo inverse equal to the number itself is 6.

Input: N = 11
Output: 10

Naive Approach: The simplest approach to solve this problem is to traverse all natural numbers from 1 to N and find the largest number such that modulo multiplicative inverse of the number under modulo N is equal to the number itself.



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

Efficient Approach: To optimize the above approach the idea is based on the following observations:

The nearest smaller number to N having modulo multiplicative inverse equal to the number itself is (N – 1).

Mathematical proof:
If X and Y are two numbers such that (X * Y) % N = 1 mod(N), then Y is modulo inverse of X.
Put X = N – 1 then
=>((N – 1) * Y) % N = 1 mod(N)
=>(N × Y) % N – Y % N = 1 mod(N)
=> Y = N – 1 
Therefore, for X = N – 1 the value of Y is equal to X.

Therefore, to solve the problem, simply print N – 1 as the required answer.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the nearest
// smaller number satisfying
// the condition
int clstNum(int N)
{
    return (N - 1);
}
 
// Driver Code
int main()
{
    int N = 11;
    cout << clstNum(N);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to find the nearest
// smaller number satisfying
// the condition
static int clstNum(int N){ return (N - 1); }
 
// Driver Code
public static void main(String[] args)
{
    int N = 11;
     
    System.out.println(clstNum(N));
}
}
 
// This code is contributed by akhilsaini

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to implement
# the above approach
 
# Function to find the nearest
# smaller number satisfying
# the condition
def clstNum(N):
  return (N - 1)
 
# Driver Code
if __name__ == '__main__':
   
  N = 11
   
  print(clstNum(N))
     
# This code is contributed by akhilsaini

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find the nearest
// smaller number satisfying
// the condition
static int clstNum(int N){ return (N - 1); }
 
// Driver Code
public static void Main()
{
    int N = 11;
     
    Console.Write(clstNum(N));
}
}
 
// This code is contributed by akhilsaini

chevron_right


Output: 

10



 

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Active and well versed member of Competitive Programming

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 : akhilsaini