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# Find three integers less than or equal to N such that their LCM is maximum

Given a number N(>=3). The task is to find the three integers (<=N) such that LCM of these three integers is maximum.
Examples:

```Input: N = 3
Output: 1 2 3

Input: N = 5
Output: 3 4 5```

Recommended Practice

Approach: Since the task is to maximize the LCM, so if all three numbers don’t have any common factor then the LCM will be the product of those three numbers and that will be maximum.

• If n is odd then the answer will be n, n-1, n-2.
• If n is even,
1. If gcd of n and n-3 is 1 then answer will be n, n-1, n-3.
2. Otherwise, n-1, n-2, n-3 will be required answer.

Below is the implementation of the above approach:

## C++

 `// CPP Program to find three integers``// less than N whose LCM is maximum``#include ``using` `namespace` `std;` `// function to find three integers``// less than N whose LCM is maximum``void` `MaxLCM(``int` `n)``{``    ``// if n is odd``    ``if` `(n % 2 != 0)``        ``cout << n << ``" "` `<< (n - 1) << ``" "` `<< (n - 2);` `    ``// if n is even and n, n-3 gcd is 1``    ``else` `if` `(__gcd(n, (n - 3)) == 1)``        ``cout << n << ``" "` `<< (n - 1) << ``" "` `<< (n - 3);` `    ``else``        ``cout << (n - 1) << ``" "` `<< (n - 2) << ``" "` `<< (n - 3);``}` `// Driver code``int` `main()``{``    ``int` `n = 12;` `    ``// function call``    ``MaxLCM(n);` `    ``return` `0;``}`

## Java

 `// Java Program to find three integers``// less than N whose LCM is maximum` `import` `java.io.*;` `class` `GFG {``   ``// Recursive function to return gcd of a and b``static` `int` `__gcd(``int` `a, ``int` `b)``{``    ``// Everything divides 0 ``    ``if` `(a == ``0``)``       ``return` `b;``    ``if` `(b == ``0``)``       ``return` `a;``   ` `    ``// base case``    ``if` `(a == b)``        ``return` `a;``   ` `    ``// a is greater``    ``if` `(a > b)``        ``return` `__gcd(a-b, b);``    ``return` `__gcd(a, b-a);``}` `// function to find three integers``// less than N whose LCM is maximum``static` `void` `MaxLCM(``int` `n)``{``    ``// if n is odd``    ``if` `(n % ``2` `!= ``0``)``        ``System.out.print(n + ``" "` `+ (n - ``1``) + ``" "` `+ (n - ``2``));` `    ``// if n is even and n, n-3 gcd is 1``    ``else` `if` `(__gcd(n, (n - ``3``)) == ``1``)``        ``System.out.print( n + ``" "` `+(n - ``1``)+ ``" "` `+ (n - ``3``));` `    ``else``        ``System.out.print((n - ``1``) + ``" "` `+ (n - ``2``) + ``" "` `+ (n - ``3``));``}` `// Driver code``public` `static` `void` `main (String[] args) {``    ``int` `n = ``12``;` `    ``// function call``    ``MaxLCM(n);``    ``}``}``// This code is contributed by anuj_67..`

## Python3

 `# Python 3 Program to find three integers``# less than N whose LCM is maximum``from` `math ``import` `gcd` `# function to find three integers``# less than N whose LCM is maximum``def` `MaxLCM(n) :` `    ``# if n is odd``    ``if` `(n ``%` `2` `!``=` `0``) :``        ``print``(n, (n ``-` `1``), (n ``-` `2``))` `    ``# if n is even and n, n-3 gcd is 1``    ``elif` `(gcd(n, (n ``-` `3``)) ``=``=` `1``) :``        ``print``(n, (n ``-` `1``), (n ``-` `3``))` `    ``else` `:``        ``print``((n ``-` `1``), (n ``-` `2``), (n ``-` `3``))` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``n ``=` `12` `    ``# function call``    ``MaxLCM(n)` `# This code is contributed by Ryuga`

## C#

 `// C# Program to find three integers``// less than N whose LCM is maximum` `using` `System;` `class` `GFG {``// Recursive function to return gcd of a and b``static` `int` `__gcd(``int` `a, ``int` `b)``{``    ``// Everything divides 0``    ``if` `(a == 0)``    ``return` `b;``    ``if` `(b == 0)``    ``return` `a;``    ` `    ``// base case``    ``if` `(a == b)``        ``return` `a;``    ` `    ``// a is greater``    ``if` `(a > b)``        ``return` `__gcd(a-b, b);``    ``return` `__gcd(a, b-a);``}` `// function to find three integers``// less than N whose LCM is maximum``static` `void` `MaxLCM(``int` `n)``{``    ``// if n is odd``    ``if` `(n % 2 != 0)``        ``Console.Write(n + ``" "` `+ (n - 1) + ``" "` `+ (n - 2));` `    ``// if n is even and n, n-3 gcd is 1``    ``else` `if` `(__gcd(n, (n - 3)) == 1)``        ``Console.Write( n + ``" "` `+(n - 1)+ ``" "` `+ (n - 3));` `    ``else``        ``Console.Write((n - 1) + ``" "` `+ (n - 2) + ``" "` `+ (n - 3));``}` `// Driver code``public` `static` `void` `Main () {``    ``int` `n = 12;` `    ``// function call``    ``MaxLCM(n);``    ``}``}``// This code is contributed by anuj_67..`

## PHP

 ` ``\$b``)``        ``return` `__gcd(``\$a` `- ``\$b``, ``\$b``);``    ``return` `__gcd(``\$a``, ``\$b` `- ``\$a``);``} ` `// function to find three integers``// less than N whose LCM is maximum``function` `MaxLCM(``\$n``)``{``    ``// if n is odd``    ``if` `(``\$n` `% 2 != 0)``        ``echo` `\$n` `, ``" "` `, (``\$n` `- 1) ,``                  ``" "` `, (``\$n` `- 2);` `    ``// if n is even and n, n-3 gcd is 1``    ``else` `if` `(__gcd(``\$n``, (``\$n` `- 3)) == 1)``        ``echo` `\$n` `, ``" "` `, (``\$n` `- 1),``                  ``" "` `, (``\$n` `- 3);`` ` `    ``else``        ``echo` `(``\$n` `- 1) , ``" "` `, (``\$n` `- 2),``                        ``" "` `, (``\$n` `- 3);``}` `// Driver code``\$n` `= 12;` `// function call``MaxLCM(``\$n``);` `// This code is contributed by Sachin``?>`

## Javascript

 ``

Output:

`11 10 9`

Time Complexity: O(log(min(a, b))), where a and b are two parameters of gcd.

Auxiliary Space: O(log(min(a, b)))

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