Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3 in C++

We are given two numbers L and R that define a range [L,R]. The goal is to find all numbers between L and R that are even, and sum of whose digits is divisible by 3.

We will do this by calculating the sum of digits of all even numbers between L and R and increment count if that sum%3==0.

Let us understand with examples.

Input − L=10, R=20

Output − Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3: 2

Explanation − Numbers between 10 and 20 that are even. 10,12,14,16,18,20. Sum of whose digits divisible by 3= 12 and 18.

Input − L=100, R=108

Output− Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3: 2

Explanation − Numbers between 100 and 108 that are even. 100,102,104,106,108. Sum of whose digits divisible by 3= 102 and 108.

Approach used in the below program is as follows

• We take variables first and last to define range.

• Function Digit_sum(int num) takes the number and returns the sum of its digits.

• Using while loop, till num!=0, add num%10, (unit digit) to total.

• Divide num by 10 to reduce it.

• At the end total will have sum of all digits.

• Function divisible_3(int first, int last) takes the range of numbers and returns the count of even numbers that have digit sum divisible by 3.

• Starting from index i=first to i<=last. Check if number i is even. (i%2==0).

• If true, then calculate sum of digits of i by calling Digit_sum(i). If that sum%3==0. Then increment count.

• At the end of for loop return count as result.

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; int Digit_sum(int num){    int total = 0;    while (num!= 0){       total += num % 10;       num = num / 10;    }    return total; } int divisible_3(int first, int last){    int count = 0;    for (int i = first; i <= last; i++){       if (i % 2 == 0 && Digit_sum(i) % 3 == 0){          count++;       }    }    return count; } int main(){    int first = 300, last = 500;    cout<<"Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3 are: "<<divisible_3(first, last);    return 0; }

Output

If we run the above code it will generate the following output −

Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3 are: 34