Java 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 5 in Fibonacci Series is 832040 
which appears at position 30.

An Efficient Solution is based on the below interesting property. The 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.

Java

// Java Program to find position of n'th multiple
// of a number k in Fibonacci Series
import java.io.*;
import java.lang.*;

public class GFG {
    public static int findPosition(int k, int n)
    {
        long f1 = 0, f2 = 1, f3;
        int i = 2;

        while (i != 0) {
            f3 = f1 + f2;
            f1 = f2;
            f2 = f3;

            if (f2 % k == 0) {
                return n * i;
            }

            i++;
        }
        return 0;
    }

    public static void main(String[] args)
    {
        // Multiple no.
        int n = 5;

        // Number of whose multiple we are finding
        int k = 4;

        System.out.print("Position of n'th multiple" +
                        " of k in Fibonacci Series is ");

        System.out.println(findPosition(k, n));
    }
}

// Code contributed by Mohit Gupta_OMG

Output:

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

Time complexity: O(1) since performing constant operations.

Auxiliary Space: O(1)

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