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APC-PRAC-038.c

Problem Statement

Print all combinations of two two-digit numbers such that the sum of digits of both numbers is equal. Example: 23 and 41 → (2+3) = 5, (4+1) = 5.

Metadata

Property Detail
Author Amit Dutta (amitdutta4255@gmail.com)
License MIT
Difficulty Beginner (index: 1 / 10)

Concepts

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  • Recursion
  • Sorting (possible)
  • Iteration

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Source Code

#include <stdio.h>

int main()
{
    printf("Combinations of two two-digit numbers such that the sum of digits of both numbers is equal:  ");
    int i, j, sum1, sum2, count = 0;
    for (i = 10; i <= 99; i++)
    {
        sum1 = (i % 10) + (i / 10);
        for (j = i + 1; j <= 99; j++)
        {
            sum2 = (j % 10) + (j / 10);
            if (sum1 == sum2)
            {
                printf("(%d, %d)  ", i, j);
                count++;
            }
        }
    }
    printf("\nCount: %d\n", count);
    return 0;
}

Explanation

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    You are explaining a C programming code to a beginner.

    STRICT RULES:

    - Only use the given code. Do NOT assume anything not present.

    - Do NOT add extra examples.

    - Keep explanation clear and short.

    - If something is unclear, say "Not clear from code".

    - Follow the exact format below. Do NOT change headings.

    FORMAT:

    [START]

    ## What it does

    (Explain the overall purpose in 1-2 sentences)

    ## Step-by-step

    (Explain how the code works in steps, simple language)

    ## Key Concepts

    (List concepts like loop, condition, function, etc.)

    ## Notes

    (Mention any limitations, errors, or assumptions)

    [END]

    CODE (APC-PRAC-038.c):

    #include <stdio.h>

    int main()
    {
        printf("Combinations of two two-digit numbers such that the sum of digits of both numbers is equal:  ");
        int i, j, sum1, sum2, count = 0;
        for (i = 10; i <= 99; i++)
        {
            sum1 = (i % 10) + (i / 10);
            for (j = i + 1; j <= 99; j++)
            {
                sum2 = (j % 10) + (j / 10);
                if (sum1 == sum2)
                {
                    printf("(%d, %d)  ", i, j);
                    count++;
                }
            }
        }
        printf("\nCount: %d\n", count);
        return 0;
    }