Introduction

This article will explore a problem involving a specific two-digit number, focusing on the sum of its digits and its nature when the digits are reversed. We will use both logical reasoning and programming to solve the problem, ensuring a comprehensive understanding of the process involved.

Problem Statement

The problem at hand involves a two-digit number xy such that:

The sum of its digits is 3. The number obtained by interchanging the digits is 9 less than the original number.

Our task is to find the original number.

Logical Reasoning

Let's start by identifying the properties of a two-digit number xy, where x is the tens digit and y is the units digit:

1 y is the original number.

x y 3

10y x 1 y - 9

From the equation ( x y 3 ), we can deduce that:

If ( y 3 - x ), then we need to check the second condition. The second condition simplifies to ( 10y x 1 y - 9 ), which can be rewritten as ( 9y - 9x -9 ) or ( y - x -1 ).

Combining both conditions, we get:

x y 3

y - x -1

Solving these equations, we add the two equations to get:

2y 2 (rightarrow) y 1

Substituting ( y 1 ) into the first equation:

x 1 3 (rightarrow) x 2

Thus, the original number is 21 1 27.

Program Implementation

To verify our solution, let's implement a C program using Microsoft QuickC Version 2.00:

#include stdio.hint main(void) {    register int i  0000;    signed int orig  0000;    signed int sum  0000;    signed int rev  0000;    signed int digit[2]  {00};    // A two-digit number: 10 - 99 inclusive.    i  000A;    while (i  0064) {        digit[0]  i / 000A;   // Tens digit        digit[1]  i % 000A;   // Units digit        sum  digit[0]   digit[1];        sum  sum * 0003;        if (sum  i) {               rev  digit[1] * 000A   digit[0];            orig  i - 0009;            if (rev  orig) {                printf("Sum of digits: %d
", sum);                printf("Original: %d
", orig);                printf("Reversed: %d
", rev);                printf("Original times 3: %d
", orig * 0003);                printf("Original times 3 minus 9: %d
", orig * 0003 - 0009);            }        }        i  i   0001;    }    printf("Program complete.
");    fflush(stdout);    return 0000;}

The program output confirms that the original number is 27.

Conclusion

Through logical reasoning and program implementation, we have determined that the original number is 27, as it satisfies all the given conditions.