# C/C++ Program for nth multiple of a number in Fibonacci Series

Given two integers n and k. Find position the n’th multiple of K in the Fibonacci series.

Examples:

```Input : k = 2, n = 3
Output : 9
3'rd multiple of 2 in Fibonacci Series is 34
which appears at position 9.

Input  : k = 4, n = 5
Output : 30
5'th multiple of 4 in Fibonacci Series is 832040
which appears at position 30.```

An Efficient Solution is based on below interesting property.
Fibonacci series is always periodic under modular representation. Below are examples.

```F (mod 2) = 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0,
1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0
Here 0 is repeating at every 3rd index and
the cycle repeats at every 3rd index.

F (mod 3) = 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, 2, 2
Here 0 is repeating at every 4th index and
the cycle repeats at every 8th index.

F (mod 4) = 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3,
1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0
Here 0 is repeating at every 6th index and
the cycle repeats at every 6th index.

F (mod 5) = 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0,
2, 2, 4, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0
Here 0 is repeating at every 5th index and
the cycle repeats at every 20th index.

F (mod 6) = 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4,
3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 2
Here 0 is repeating at every 12th index and
the cycle repeats at every 24th index.

F (mod 7) = 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1,
0, 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6
Here 0 is repeating at every 8th index and
the cycle repeats at every 16th index.

F (mod 8) = 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2,
3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0
Here 0 is repeating at every 6th index and
the cycle repeats at every 12th index.

F (mod 9) = 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7,
6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8
Here 0 is repeating at every 12th index and
the cycle repeats at every 24th index.

F (mod 10) = 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0,
7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0.
Here 0 is repeating at every 15th index and
the cycle repeats at every 60th index.```

## C++

 `// C++ program to find position of n'th multiple``// of a number k in Fibonacci Series``#include ``using` `namespace` `std;``const` `int` `MAX = 1000;` `// Returns position of n'th multiple of k in``// Fibonacci Series``int` `findPosition(``int` `k, ``int` `n)``{``    ``// Iterate through all fibonacci numbers``    ``unsigned ``long` `long` `int` `f1 = 0, f2 = 1, f3;``    ``for` `(``int` `i = 2; i <= MAX; i++) {``        ``f3 = f1 + f2;``        ``f1 = f2;``        ``f2 = f3;` `        ``// Found first multiple of k at position i``        ``if` `(f2 % k == 0)` `            ``// n'th multiple would be at position n*i``            ``// using Periodic property of Fibonacci``            ``// numbers under modulo.``            ``return` `n * i;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `n = 5, k = 4;``    ``cout << ``"Position of n'th multiple of k"``         ``<< ``" in Fibonacci Series is "``         ``<< findPosition(k, n) << endl;``    ``return` `0;``}`

Output:
`Position of n'th multiple of k in Fibonacci Series is 30`

Time Complexity: O(1000), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on n’th multiple of a number in Fibonacci Series for more details!

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