# Sum of integers upto N with given unit digit

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: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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



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 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



Output:

39

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