# 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 ``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

 ``
Output:
`10`

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

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