# Program to find LCM of two Fibonnaci Numbers

Given here are two positive numbers a and b. The task is to print the least common multiple of a’th and b’th Fibonacci Numbers.

The first few Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ……

Note that 0 is considered as 0’th Fibonacci Number.

Examples:

```Input : a = 3, b = 12
Output : 144

Input : a = 8, b = 37
Output : 507314157
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The simple solution of the problem is,

1. Find the a’th fibonacci number.
2. Find the b’th fibonacci number.
3. Find their GCD, and with the help of the GCD find their LCM. The relation is LCM(a, b) = (a x b) / GCD(a, b) (Please refer here).

Below is the implementation of the above approach:

## C++

 `// C++ Program to find LCM of Fib(a) ` `// and Fib(b) ` `#include ` `using` `namespace` `std; ` `const` `int` `MAX = 1000; ` ` `  `// Create an array for memoization ` `int` `f[MAX] = { 0 }; ` ` `  `// Function to return the n'th Fibonacci ` `// number using table f[]. ` `int` `fib(``int` `n) ` `{ ` `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return` `0; ` `    ``if` `(n == 1 || n == 2) ` `        ``return` `(f[n] = 1); ` ` `  `    ``// If fib(n) is already computed ` `    ``if` `(f[n]) ` `        ``return` `f[n]; ` ` `  `    ``int` `k = (n & 1) ? (n + 1) / 2 : n / 2; ` ` `  `    ``// Applying recursive formula ` `    ``// Note value n&1 is 1 ` `    ``// if n is odd, else 0. ` `    ``f[n] = (n & 1) ? (fib(k) * fib(k) + fib(k - 1) * fib(k - 1)) ` `                   ``: (2 * fib(k - 1) + fib(k)) * fib(k); ` ` `  `    ``return` `f[n]; ` `} ` ` `  `// Function to return gcd of a and b ` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(a == 0) ` `        ``return` `b; ` ` `  `    ``return` `gcd(b % a, a); ` `} ` ` `  `// Function to return the LCM of ` `// Fib(a) and Fib(a) ` `int` `findLCMFibonacci(``int` `a, ``int` `b) ` `{ ` `    ``return` `(fib(a) * fib(b)) / fib(gcd(a, b)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a = 3, b = 12; ` ` `  `    ``cout << findLCMFibonacci(a, b); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java ram to find LCM of Fib(a) ` `// and Fib(b) ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `MAX = ``1000``; ` ` `  `// Create an array for memoization ` `static` `int``[] f = ``new` `int``[MAX]; ` ` `  `// Function to return the n'th Fibonacci ` `// number using table f[]. ` `static` `int` `fib(``int` `n) ` `{ ` `    ``// Base cases ` `    ``if` `(n == ``0``) ` `        ``return` `0``; ` `    ``if` `(n == ``1` `|| n == ``2``) ` `        ``return` `(f[n] = ``1``); ` ` `  `    ``// If fib(n) is already computed ` `    ``if` `(f[n] != ``0``) ` `        ``return` `f[n]; ` `    ``int` `k = ``0``;  ` `    ``if` `((n & ``1``) != ``0``) ` `        ``k = (n + ``1``) / ``2``; ` `    ``else` `        ``k = n / ``2``; ` ` `  `    ``// Applying recursive formula ` `    ``// Note value n&1 is 1 ` `    ``// if n is odd, else 0. ` `    ``if``((n & ``1` `) != ``0``) ` `        ``f[n] = (fib(k) * fib(k) +  ` `                ``fib(k - ``1``) * fib(k - ``1``)); ` `    ``else` `        ``f[n] = (``2` `* fib(k - ``1``) +  ` `                    ``fib(k)) * fib(k); ` ` `  `    ``return` `f[n]; ` `} ` ` `  `// Function to return gcd of a and b ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(a == ``0``) ` `        ``return` `b; ` ` `  `    ``return` `gcd(b % a, a); ` `} ` ` `  `// Function to return the LCM of ` `// Fib(a) and Fib(a) ` `static` `int` `findLCMFibonacci(``int` `a, ``int` `b) ` `{ ` `    ``return` `(fib(a) * fib(b)) / fib(gcd(a, b)); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `a = ``3``, b = ``12``; ` ` `  `    ``System.out.println(findLCMFibonacci(a, b)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Surendra_Gangwar `

## Python3

 `# Python 3 Program to find LCM of  ` `# Fib(a) and Fib(b) ` `MAX` `=` `1000` ` `  `# Create an array for memoization ` `f ``=` `[``0``] ``*` `MAX` ` `  `# Function to return the n'th  ` `# Fibonacci number using table f[]. ` `def` `fib(n): ` ` `  `    ``# Base cases ` `    ``if` `(n ``=``=` `0``): ` `        ``return` `0` `    ``if` `(n ``=``=` `1` `or` `n ``=``=` `2``): ` `        ``f[n] ``=` `1` `        ``return` `f[n] ` ` `  `    ``# If fib(n) is already computed ` `    ``if` `(f[n]): ` `        ``return` `f[n] ` ` `  `    ``k ``=` `(n ``+` `1``) ``/``/` `2` `if` `(n & ``1``) ``else` `n ``/``/` `2` ` `  `    ``# Applying recursive formula ` `    ``# Note value n&1 is 1 ` `    ``# if n is odd, else 0. ` `    ``if` `(n & ``1``): ` `        ``f[n] ``=` `(fib(k) ``*` `fib(k) ``+`  `                ``fib(k ``-` `1``) ``*` `fib(k ``-` `1``)) ` `    ``else``: ` `        ``f[n] ``=` `(``2` `*` `fib(k ``-` `1``) ``+` `fib(k)) ``*` `fib(k) ` ` `  `    ``return` `f[n] ` ` `  `# Function to return gcd of a and b ` `def` `gcd(a, b): ` `    ``if` `(a ``=``=` `0``): ` `        ``return` `b ` ` `  `    ``return` `gcd(b ``%` `a, a) ` ` `  `# Function to return the LCM of ` `# Fib(a) and Fib(a) ` `def` `findLCMFibonacci(a, b): ` ` `  `    ``return` `(fib(a) ``*` `fib(b)) ``/``/` `fib(gcd(a, b)) ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``a ``=` `3` `    ``b ``=` `12` ` `  `    ``print` `(findLCMFibonacci(a, b)) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# ram to find LCM of Fib(a) ` `// and Fib(b) ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `static` `int` `MAX = 1000; ` ` `  `// Create an array for memoization ` `static` `int``[] f = ``new` `int``[MAX]; ` ` `  `// Function to return the n'th Fibonacci ` `// number using table f[]. ` `static` `int` `fib(``int` `n) ` `{ ` `    ``// Base cases ` `    ``if` `(n == 0) ` `        ``return` `0; ` `    ``if` `(n == 1 || n == 2) ` `        ``return` `(f[n] = 1); ` ` `  `    ``// If fib(n) is already computed ` `    ``if` `(f[n] != 0) ` `        ``return` `f[n]; ` `    ``int` `k = 0;  ` `    ``if` `((n & 1) != 0) ` `        ``k = (n + 1) / 2; ` `    ``else` `        ``k = n / 2; ` ` `  `    ``// Applying recursive formula ` `    ``// Note value n&1 is 1 ` `    ``// if n is odd, else 0. ` `    ``if``((n & 1 ) != 0) ` `        ``f[n] = (fib(k) * fib(k) +  ` `                ``fib(k - 1) * fib(k - 1)); ` `    ``else` `        ``f[n] = (2 * fib(k - 1) +  ` `                    ``fib(k)) * fib(k); ` ` `  `    ``return` `f[n]; ` `} ` ` `  `// Function to return gcd of a and b ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(a == 0) ` `        ``return` `b; ` ` `  `    ``return` `gcd(b % a, a); ` `} ` ` `  `// Function to return the LCM of ` `// Fib(a) and Fib(a) ` `static` `int` `findLCMFibonacci(``int` `a, ``int` `b) ` `{ ` `    ``return` `(fib(a) * fib(b)) / fib(gcd(a, b)); ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `a = 3, b = 12; ` ` `  `    ``Console.WriteLine(findLCMFibonacci(a, b)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## PHP

 ` `

Output:

```144
```

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