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Sum of integers upto N with given unit digit
  • Last Updated : 20 Apr, 2021

Given two integer N and D, the task is to find the sum of all the integers from 1 to N whose unit digit is D.
Examples: 
 

Input: N = 30, D = 3 
Output: 39 
3 + 13 + 23 = 39
Input: N = 5, D = 7 
Output:
 

 

Naive approach: 
 

  • Traverse from 1 to N.
  • If the unit digit of the number is D add the number to the sum.
  • Finally print the value of sum.

Below is the implementation of the above approach:
 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the required sum
ll getSum(int n, int d)
{
    ll sum = 0;
    for (int i = 1; i <= n; i++) {
 
        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}
 
// Driver code
int main()
{
    int n = 30, d = 3;
    cout << getSum(n, d);
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class solution
{
 
// Function to return the required sum
static long getSum(int n, int d)
{
    long sum = 0;
    for (int i = 1; i <= n; i++) {
 
        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}
 
// Driver code
public static void main(String args[])
{
    int n = 30, d = 3;
    System.out.println(getSum(n, d));
    
}
}

Python3




# Python3 implementation of the approach
 
# Function to return the required sum
def getSum(n, d) :
    sum = 0;
    for i in range(n + 1) :
 
        # If the unit digit is d
        if (i % 10 == d) :
            sum += i
    return sum
 
# Driver code
if __name__ == "__main__" :
 
    n , d = 30, 3
    print(getSum(n, d))
 
# This code is contributed by Ryuga

C# 


// C# implementation of the approach
using System;
class gfg
{
  // Function to return the required sum
 public static int getSum(int n, int d)
 {
    int sum = 0;
    for (int i = 1; i <= n; i++) {

        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}

// Driver code
 public static int Main()
 { 
    int n = 30, d = 3;
    Console.WriteLine( getSum(n, d));
    return 0;
 }
}


PHP




<?php
// PHP implementation of the approach
 
// Function to return the required sum
function getSum($n, $d)
{
    $sum = 0;
    for ($i = 1; $i <= $n; $i++)
    {
 
        // If the unit digit is d
        if ($i % 10 == $d)
            $sum += $i;
    }
    return $sum;
}
 
// Driver code
$n = 30;
$d = 3;
echo getSum($n, $d);
     
// This code is contributed by Sachin
?>

Javascript




<script>
 
// java script implementation of the approach
 
// Function to return the required sum
function getSum(n, d)
{
    let sum = 0;
    for (let i = 1; i <= n; i++)
    {
 
        // If the unit digit is d
        if (i % 10 == d)
            sum += i;
    }
    return sum;
}
 
// Driver code
let n = 30;
let d = 3;
document.write( getSum(n, d));
 
// This code is contributed
// by bobby
</script>
Output: 
39

 

Efficient approach: While D < N update sum = sum + D and D = D + 10. Print the sum in the end.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to return the required sum
ll getSum(int n, int d)
{
    ll sum = 0;
    while (d <= n)
    {
        sum += d;
        d += 10;
    }
    return sum;
}
 
// Driver code
int main()
{
    int n = 30, d = 3;
    cout << getSum(n, d);
    return 0;
}

Java




// Java implementation of the approach
class Solution
{
 
  
// Function to return the required sum
static long getSum(int n, int d)
{
    long sum = 0;
    while (d <= n) {
        sum += d;
        d += 10;
    }
    return sum;
}
  
// Driver code
public static void main(String args[])
{
    int n = 30, d = 3;
    System.out.print(getSum(n, d));
     
}
}
//contributed by Arnab Kundu

Python3




# Python3 implementation of the approach
# Function to return the required sum
def getSum(n, d):
    sum = 0
    while (d <= n):
        sum += d
        d += 10
    return sum
 
# Driver code
n = 30
d = 3
print(getSum(n, d))
 
# This code is contributed
# by sahishelangia

C#




// C# implementation of the approach
using System;
class GFG
{
 
// Function to return the required sum
static long getSum(int n, int d)
{
    long sum = 0;
    while (d <= n)
    {
        sum += d;
        d += 10;
    }
    return sum;
}
 
// Driver code
public static void Main()
{
    int n = 30, d = 3;
    Console.Write(getSum(n, d));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// PHP implementation of the approach
// Function to return the required sum
function getSum($n, $d)
{
    $sum = 0;
    while ($d <= $n)
    {
        $sum += $d;
        $d += 10;
    }
    return $sum;
}
 
// Driver code
$n = 30; $d = 3;
echo(getSum($n, $d));
     
// This Code is contributed
// by Mukul Singh
?>

Javascript




<script>
// java script  implementation of the approach
// Function to return the required sum
function getSum(n, d)
{
    let sum = 0;
    while (d <= n)
    {
        sum += d;
        d += 10;
    }
    return sum;
}
 
// Driver code
let n = 30;
let d = 3;
document.write(getSum(n, d));
    
// This code is contributed
// by sravan kumar
</script>
Output: 
39

 

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