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.

Fibonacci Series(F) : 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040… (neglecting the first 0).

A **Simple Solution** is to traverse Fibonacci numbers starting from first number. While traversing, keep track of counts of multiples of k. Whenever the count becomes n, return the position.

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.

**Why is Fibonacci Series Periodic under Modulo?**

Under modular representation, we know that each Fibonacci number will be represented as some residue 0 ? F (mod m) < m. Thus, there are only m possible values for any given F (mod m) and hence m*m = m^2 possible pairs of consecutive terms within the sequence. Since m^2 is ?nite, we know that some pair of terms must eventually repeat itself. Also, as any pair of terms in the Fibonacci sequence determines the rest of the sequence, we see that the Fibonacci series modulo m must repeat itself at some point, and thus must be periodic.
Source : https://www.whitman.edu/Documents/Academics/Mathematics/clancy.pdf

Based on above fact, we can quickly find position of n’th multiple of K by simply finding first multiple. If position of first multiple is i, we return position as n*i.

Below is C++ implementation.

## C/C++

// C++ program to find position of n'th multiple // of a mumber k in Fibonacci Series # include <bits/stdc++.h> using namespace std; const int MAX = 1000; // Returns position of n'th multple 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; }

## Java

// Java Program to find position of n'th multiple // of a mumber k in Fibonacci Series package gfg; 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

## Python3

# Python Program to find position of n'th multiple # of a mumber k in Fibonacci Series def findPosition(k, n): f1 = 0 f2 = 1 i =2; while i!=0: f3 = f1 + f2; f1 = f2; f2 = f3; if f2%k == 0: return n*i i+=1 return # Multiple no. n = 5; # Number of whose multiple we are finding k = 4; print("Position of n'th multiple of k in" "Fibonacci Seires is", findPosition(k,n)); # Code contributed by Mohit Gupta_OMG

**Output :**

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

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