Numbers Divisible by 7 in C – List, Count, and Sum

Finding all integers in a range that are divisible by a given number is a classic loop exercise. The key tool is the modulo operator % — if i % 7 == 0, then i is exactly divisible by 7 with no remainder. This post shows two programs: one for the classic 100–200 range, and a flexible version where the user supplies the range and divisor.

Method 1 — Numbers from 100 to 200 Divisible by 7

#include <stdio.h>

int main(void)
{
    int i, count = 0, sum = 0;

    printf("Numbers from 100 to 200 divisible by 7:\n");

    for (i = 100; i <= 200; i++) {
        if (i % 7 == 0) {
            printf("%d ", i);
            sum += i;
            count++;
        }
    }

    printf("\n\nCount : %d\n", count);
    printf("Sum   : %d\n", sum);

    return 0;
}

Output

Numbers from 100 to 200 divisible by 7:
105 112 119 126 133 140 147 154 161 168 175 182 189 196

Count : 14
Sum   : 2107

Method 2 — User-Supplied Range and Divisor

The same logic works for any range and any divisor. This version asks the user for all three values and guards against division by zero.

#include <stdio.h>

int main(void)
{
    int start, end, divisor, i, count = 0, sum = 0;

    printf("Enter start, end, and divisor: ");
    scanf("%d %d %d", &start, &end, &divisor);

    if (divisor == 0) {
        printf("Error: divisor cannot be zero.\n");
        return 1;
    }

    printf("\nNumbers from %d to %d divisible by %d:\n",
           start, end, divisor);

    for (i = start; i <= end; i++) {
        if (i % divisor == 0) {
            printf("%d ", i);
            sum += i;
            count++;
        }
    }

    if (count == 0)
        printf("(none)");

    printf("\n\nCount : %d\n", count);
    printf("Sum   : %d\n", sum);

    return 0;
}

More Sample Runs

Enter start, end, and divisor: 1 50 5

Numbers from 1 to 50 divisible by 5:
5 10 15 20 25 30 35 40 45 50

Count : 10
Sum   : 275
Enter start, end, and divisor: 1 10 11

Numbers from 1 to 10 divisible by 11:
(none)

Count : 0
Sum   : 0

How to Compile and Run

gcc -ansi -Wall -Wextra -o divisible divisible.c
./divisible

Code Explanation

The modulo operator % returns the remainder after integer division. 105 % 7 is 0 because 105 = 7 × 15 exactly. 106 % 7 is 1 because 106 = 7 × 15 + 1. So i % 7 == 0 is the precise test for divisibility.

Why the zero check? The modulo operator with a zero right-hand operand is undefined behaviour in C — it corresponds to dividing by zero. Method 2 checks for this before entering the loop and exits with an error code of 1, which signals failure to the shell.

Counting vs listing: count++ inside the if block increments a counter each time a match is found. sum += i accumulates the running total. Both finish in a single pass through the loop — no second iteration needed.

What This Program Teaches

  • How the modulo operator % tests for exact divisibility
  • How to count and sum matching values inside a loop with a single pass
  • How to handle an invalid input (zero divisor) with an early return
  • How to generalise a hardcoded solution into a user-driven one

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