Let the original number be $\overline{xy}$, where $x$ is the digit at the tens place and $y$ is the digit at the units place. Thus, the number can be written as $10x + y$.

According to the problem, we have two conditions:

1. The digit at the tens place (x) is 3 more than two times the digit at the units place (y).
2. The number formed by reversing the digits (i.e., $\overline{yx}$ or $10y + x$) is 8 more than one third of the original number.

We can express these conditions mathematically as follows:

1. $x = 2y + 3$
2. $10y + x = \frac{1}{3}(10x + y) + 8$

Let's solve these equations step by step.

Step 1: Solve for $x$ in terms of $y$

From the first condition: $x = 2y + 3$

Step 2: Substitute $x$ in the second condition

Substitute $x = 2y + 3$ into the second equation: $10y + (2y + 3) = \frac{1}{3}(10(2y + 3) + y) + 8$

Simplify the left side: $10y + 2y + 3 = 12y + 3$

Simplify the right side: $\frac{1}{3}(20y + 30 + y) + 8 = \frac{1}{3}(21y + 30) + 8 = 7y + 10 + 8 = 7y + 18$

So we have: $12y + 3 = 7y + 18$

Step 3: Solve for $y$

$12y + 3 = 7y + 18$ Subtract $7y$ from both sides: $5y + 3 = 18$ Subtract 3 from both sides: $5y = 15$ Divide by 5: $y = 3$

Step 4: Find $x$

$x = 2y + 3$ Substitute $y = 3$: $x = 2(3) + 3 = 6 + 3 = 9$

Step 5: Determine the original number

The original number is $\overline{xy} = 93$.

Verification

1. The digit at the tens place (9) is 3 more than two times the digit at the units place (3): $9 = 2(3) + 3 = 6 + 3 = 9$ (Correct)

2. The number formed by reversing the digits (39) is 8 more than one third of the original number (93): $10y + x = 39$ $\frac{1}{3}(93) + 8 = 31 + 8 = 39$ (Correct)

Thus, the original number is $\boxed{93}$.

Would you like more details or have any questions about this solution? Here are some related questions you might find interesting:

1. What if the tens digit was 4 more than two times the units digit?
2. How do you solve similar problems involving three-digit numbers?
3. How does changing the relationship between digits affect the solution?
4. What if the reversed number was less than one third of the original number?
5. How do such digit problems relate to algebraic equations?
6. Can we generalize this method to solve problems with more complex conditions?
7. How do these types of problems help in understanding number properties?
8. How can we use these problems to enhance logical reasoning skills?

Tip: Always double-check your calculations by substituting the values back into the original conditions.