C program to calculate sum of first and last digit values is equal to product of remaining two digits?

Question

Input: 1223
1+3=2*2
that means if sum of 1st and last digit is equal to product of remaining 2 digits then print 1+3=2*2 else error msg.

Answer 1

Here is my program(Tested).

#include <stdio.h>

int main()
{
    int num = 0;
    int n[4], i = 3;

    printf("Enter a 4 digit number : ");
    scanf("%d", &num);

    //I am asumming that, you have enterd a 4 digit number.
    while(i >= 0 ) {
        n[i--] = num % 10;
        num /= 10;
    }

    if ((n[0] + n[3]) == (n[1] * n[2])) {
        printf("%d + %d = %d * %d\n", n[0], n[3], n[1], n[2]);
    } else {
        printf("Error : (Sum of 1st and last is not equal to porduct of remaing  two).\n");
    }

    return 0;
}

Comments

Arshad bhai without taking array variable can't we solve this problem i f we can solve the problem pls mail me the program

---

Yes  absolutely, we can solve the problem without using array.
I am posting that as a second ans.

Answer 2

Second ans without using array.

#include <stdio.h>

int main()
{
    int num = 0;
    int sumof_first_last, prodof_rem2;

    printf("Enter a 4 digit number : ");
    scanf("%d", &num);

    //I am asumming you have enterd a 4 digit number.

    sumof_first_last = (num / 1000) + (num % 10);  // it will give first and last digit
    prodof_rem2 = ((num / 100) % 10) * ((num / 10) % 10); // it will give 2nd and 3rd digit i.e remaning two.

    if (sumof_first_last == prodof_rem2) {
        printf("Sum of 1st and last is not equal to porduct of remaing  two i.e = %d\n", sumof_first_last);
    }
    else {
        printf("Error : (Sum of 1st and last is not equal to porduct of remaing  two.\n");
    }
    return 0;
}