Minimum possible value of (i * j) % 2019

Given two integers L and R, the task is to find the minimum possible value of (i * j) % 2019 where L ≤ i < j ≤ R

Examples:

Input: L = 2020, R = 2040
Output: 2
(2020*2021)%2019 = 2



Input: L = 3, R = 4
Output: 12

Approach:

  • If R – L ≥ 2019 then answer is 0 as we will get a number in the range which is divisible by 2019 which gives remainder 0.
  • If R – L < 2019 then we can run nested loops and find the minimum value.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int MOD = 2019;
  
// Function to return the minimum
// possible value of (i * j) % 2019
int min_modulo(int l, int r)
{
    // If we can get a number
    // divisible by 2019
    if (r - l >= MOD)
        return 0;
    else {
  
        // Find the minimum value
        // by running nested loops
        int ans = MOD - 1;
        for (int i = l; i <= r; i++) {
            for (int j = i + 1; j <= r; j++) {
                ans = min(ans, (i * j) % MOD);
            }
        }
        return ans;
    }
}
  
// Driver code
int main()
{
    int l = 2020, r = 2040;
  
    cout << min_modulo(l, r);
  
    return 0;
}

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Java














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// Java implementation of the approach 


class GFG 


{ 


static int MOD = 2019; 


  


// Function to return the minimum 


// possible value of (i * j) % 2019 


static int min_modulo(int l, int r) 


{ 


    // If we can get a number 


    // divisible by 2019 


    if (r - l >= MOD) 


        return 0; 


    else 


    { 


  


        // Find the minimum value 


        // by running nested loops 


        int ans = MOD - 1; 


        for (int i = l; i <= r; i++)  


        { 


            for (int j = i + 1; j <= r; j++) 


            { 


                ans = Math.min(ans, (i * j) % MOD); 


            } 


        } 


        return ans; 


    } 


} 


  


// Driver code 


public static void main(String []args)  


{ 


    int l = 2020, r = 2040; 


  


    System.out.println(min_modulo(l, r)); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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Python3














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# Python3 implementation of the approach  


MOD = 2019;  


  


# Function to return the minimum  


# possible value of (i * j) % 2019  


def min_modulo(l, r) : 


  


    # If we can get a number  


    # divisible by 2019  


    if (r - l >= MOD) : 


        return 0;  


    else : 


  


        # Find the minimum value  


        # by running nested loops  


        ans = MOD - 1;  


        for i in range(l, r + 1) : 


            for j in range(i + 1, r + 1) : 


                ans = min(ans, (i * j) % MOD);  


          


        return ans;  


  


# Driver code  


if __name__ == "__main__" : 


      


    l = 2020; r = 2040; 


      


    print(min_modulo(l, r));  


  


# This code is contributed by AnkitRai01 



















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C#














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// C# implementation of the approach 


using System; 


  


class GFG  


{ 


static int MOD = 2019; 


  


// Function to return the minimum 


// possible value of (i * j) % 2019 


static int min_modulo(int l, int r) 


{ 


    // If we can get a number 


    // divisible by 2019 


    if (r - l >= MOD) 


        return 0; 


    else


    { 


  


        // Find the minimum value 


        // by running nested loops 


        int ans = MOD - 1; 


        for (int i = l; i <= r; i++)  


        { 


            for (int j = i + 1; j <= r; j++) 


            { 


                ans = Math.Min(ans, (i * j) % MOD); 


            } 


        } 


        return ans; 


    } 


} 


  


// Driver code 


public static void Main(String []args)  


{ 


    int l = 2020, r = 2040; 


  


    Console.WriteLine(min_modulo(l, r)); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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Output:


2







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