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 <bits/stdc++.h>
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!