Problem:
The sum of a 2-digit number and the number obtained by reversing the order of its digits is 99. If the digits of the number differ by 3, what is the number?

Solution:

To solve this problem, let's break it down into smaller steps:

Step 1: Understand the problem
We are given that the sum of a 2-digit number and the number obtained by reversing the order of its digits is 99. Additionally, the digits of the number differ by 3. We need to find the number.

Step 2: Assign variables
Let's represent the tens digit of the number as "x" and the units digit as "y". So the original number can be represented as 10x + y.

Step 3: Set up equations
We are given two conditions:
1. The sum of the original number and its reverse is 99: (10x + y) + (10y + x) = 99
2. The digits of the number differ by 3: x - y = 3

Step 4: Simplify the equations
Let's simplify the equation obtained in step 3:

(10x + y) + (10y + x) = 99
11x + 11y = 99
x + y = 9

We can also rewrite the second equation as:
x = y + 3

Step 5: Solve the system of equations
Now we have a system of two equations with two variables. We can solve this system by substitution or elimination.

Substituting the value of x from the second equation into the first equation:
(y + 3) + y = 9
2y + 3 = 9
2y = 6
y = 3

Substituting the value of y into the second equation:
x = 3 + 3
x = 6

So the original number is 63.

Step 6: Verify the solution
To verify our solution, we can check if the sum of the original number (63) and its reverse (36) is indeed 99:
63 + 36 = 99

The sum is indeed 99, so our solution is correct.

Final Answer:
The number is 63.

36 or 63