Check if a Number is Divisible by 2 and 3 in C

February 24, 2026

C Program to Solve Two Sum Using Brute Force (With Algorithm & Output)

Introduction The Two Sum problem is a popular coding interview question where we must find two indices of an array whose values add up to a given target. This program demonstrates a simple brute-force solution in C using nested loops and dynamic memory allocation. Problem Statement Given an integer array and a target value, return the indices of the two numbers such that they add up to the target. Each input has exactly one solution, and the same element cannot be used twice. The result should return the indices, not the values. If no solution exists, return NULL. Algorithm / Logic Explanation Start the program. Traverse the array using a loop from index 0 to numsSize - 1 . Inside this loop, use another loop starting from i + 1 to numsSize - 1 . For every pair (i, j) , check if nums[i] + nums[j] == target . If condition becomes true: Allocate memory for 2 integers using malloc() . Store indices i and j . Set returnSize = 2 . Return the result poi...

✅ CHECK IF A NUMBER IS DIVISIBLE BY 2 AND 3 IN C

#include <stdio.h>

int main() {
    int num;

    printf("Enter a number:\n");
    scanf("%d", &num);

    if (num % 2 == 0 && num % 3 == 0) {
        printf("%d is perfectly divisible by both 2 and 3.\n", num);
    } else {
        printf("%d is NOT divisible by both 2 and 3.\n", num);
    }

    return 0;
}

🧠 Explanation:

This program checks whether a given number is divisible by both 2 and 3.

It uses num % 2 == 0 to check divisibility by 2.
And num % 3 == 0 to check divisibility by 3.
If both are true, it prints that the number is divisible by both.

🖥️ Sample Output:

Enter a number:
12
12 is perfectly divisible by both 2 and 3.

🔑 Keywords:

C program to check divisibility, modulus operator, if condition, logical AND, basic C program.

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C Program to Solve Two Sum Using Brute Force (With Algorithm & Output)