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Nearest smaller number to N having multiplicative inverse under modulo N equal to that number
  • Last Updated : 13 Apr, 2021

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++




// 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);
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
// Javascript program to implement
// the above approach
 
// Function to find the nearest
// smaller number satisfying
// the condition
function clstNum(N)
{
    return (N - 1);
}
 
// Driver Code
let N = 11;
document.write(clstNum(N));
 
// This code is contributed by subham348.
</script>
Output: 
10

 

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

 

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