Last digit of sum of numbers in the given range in the Fibonacci series

Given two non-negative integers M, N which signifies the range [M, N] where M ≤ N, the task is to find the last digit of the sum of FM + FM+1… + FN where FK is the Kth Fibonacci number in the Fibonacci series.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …

Examples:

Input: M = 3, N = 9
Output: 6
Explanation:
We need to find F3 + F4 + F5 + F6 + F7 + F8 + F9
=> 2 + 3 + 5 + 8 + 13 + 21 + 34 = 86.
Clearly, the last digit of the sum is 6.

Input: M = 3, N = 7
Output: 1
Explanation:
We need to find F3 + F4 + F5 + F6 + F7
=> 2 + 3 + 5 + 8 + 13 = 31.
Clearly, the last digit of the sum is 1.



Naive Approach: The naive approach for this problem is to one by one find the sum of all Kth Fibonacci Numbers where K lies in the range [M, N] and return the last digit of the sum in the end. The time complexity for this approach is O(N) and this method fails for higher-ordered values of N.

Efficient Approach: An efficient approach for this problem is to use the concept of Pisano Period.

  • The idea is to calculate the sum of (M – 1) and N Fibonacci numbers respectively, and subtracting the last digit of the computed values.
  • This is because the last digit of the sum of all the Kth Fibonacci numbers such that K lies in the range [M, N] is equal to the difference of the last digits of the sum of all the Kth Fibonacci numbers in the range [0, N] and the sum of all the Kth Fibonacci numbers in the range [0, M – 1].
  • These values can respectively be calculated by the concept of the Pisano period in a very short time.
  • Let’s understand how the Pisano period works. The following table illustrates the first 10 Fibonacci numbers along with its values obtained when modulo 2 is performed on the numbers.
    i 0 1 2 3 4 5 6 7 8 9 10
    Fi 0 1 1 2 3 5 8 13 21 34 55
    Fi mod 2 0 1 1 0 1 1 0 1 1 0 10
  • Clearly, the Pisano period for (Fi mod 2) is 3 since 011 repeat itself and length(011) = 3.
  • Now, lets observe the following identity:

    7 = 2 * 3 + 1
    Dividend = (Quotient × Divisor) + Remainder
    => F7 mod 2 = F1 mod 2 = 1.

  • Therefore, instead of calculating the last digit of the sum of all numbers in the range [0, N], we simply calculate the sum until the remainder given that the Pisano period for Fi mod 10 is 60.

Below is the implementation of the above approach:

C++

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// C++ program to calculate 
// last digit of the sum of the 
// fibonacci numbers from M to N
#include<bits/stdc++.h>
using namespace std;
  
// Calculate the sum of the first 
// N Fibonacci numbers using Pisano
// period 
long long fib(long long n)
{
      
    // The first two Fibonacci numbers
    long long f0 = 0;
    long long f1 = 1;
  
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
        // Pisano period for % 10 is 60
        long long rem = n % 60;
  
        // Checking the remainder
        if(rem == 0)
           return 0;
  
        // The loop will range from 2 to 
        // two terms after the remainder
        for(long long i = 2; i < rem + 3; i++)
        {
           long long f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
          
        long long s = f1 - 1;
        return s;
    }
}
  
// Driver Code
int main()
{
    long long m = 10087887; 
    long long n = 2983097899;
  
    long long final = abs(fib(n) - fib(m - 1));
    cout << final % 10 << endl;
}
  
// This code is contributed by Bhupendra_Singh

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Java

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// Java program to calculate 
// last digit of the sum of the 
// fibonacci numbers from M to N
import java.util.*;
  
class GFG{
  
// Calculate the sum of the first 
// N Fibonacci numbers using Pisano
// period 
static int fib(long n)
{
      
    // The first two Fibonacci numbers
    int f0 = 0;
    int f1 = 1;
  
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
          
        // Pisano period for % 10 is 60
        int rem = (int) (n % 60);
  
        // Checking the remainder
        if(rem == 0)
        return 0;
  
        // The loop will range from 2 to 
        // two terms after the remainder
        for(int i = 2; i < rem + 3; i++)
        {
           int f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
          
        int s = f1 - 1;
        return s;
    }
}
  
// Driver Code
public static void main(String args[])
{
    int m = 10087887
    long n = 2983097899L;
    int Final = (int)Math.abs(fib(n) - 
                              fib(m - 1));
      
    System.out.println(Final % 10);
}
}
  
// This code is contributed by AbhiThakur

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Python3

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# Python3 program to calculate 
# Last Digit of the sum of the 
# Fibonacci numbers from M to N
  
# Calculate the sum of the first 
# N Fibonacci numbers using Pisano
# period 
def fib(n):
  
    # The first two Fibonacci numbers
    f0 = 0
    f1 = 1
  
    # Base case
    if (n == 0):
        return 0
    if (n == 1):
        return 1
    else:
  
        # Pisano Period for % 10 is 60
        rem = n % 60
  
        # Checking the remainder
        if(rem == 0):
            return 0
  
        # The loop will range from 2 to 
        # two terms after the remainder
        for i in range(2, rem + 3):
            f =(f0 + f1)% 60
            f0 = f1
            f1 = f
  
        s = f1-1
        return(s)
  
# Driver code
if __name__ == '__main__':
      
    m = 10087887 
    n = 2983097899
  
    final = fib(n)-fib(m-1)
  
    print(final % 10)

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C#

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// C# program to calculate 
// last digit of the sum of the 
// fibonacci numbers from M to N
using System;
  
class GFG{
  
// Calculate the sum of the first 
// N fibonacci numbers using Pisano
// period 
static int fib(long n)
{
      
    // The first two fibonacci numbers
    int f0 = 0;
    int f1 = 1;
  
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
          
        // Pisano period for % 10 is 60
        int rem = (int)(n % 60);
  
        // Checking the remainder
        if(rem == 0)
           return 0;
  
        // The loop will range from 2 to 
        // two terms after the remainder
        for(int i = 2; i < rem + 3; i++)
        {
           int f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
          
        int s = f1 - 1;
        return s;
    }
}
  
// Driver Code
public static void Main()
{
    int m = 10087887; 
    long n = 2983097899L;
    int Final = (int)Math.Abs(fib(n) - 
                              fib(m - 1));
      
    Console.WriteLine(Final % 10);
}
}
  
// This code is contributed by Code_Mech

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Output:

5

Time Complexity: O(1), because this code runs almost 60 times for any input number.

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