Let's break down the problem step by step:

1. Let the three-digit number be represented as $100h + 10t + u$, where $h$ is the hundreds digit, $t$ is the tens digit, and $u$ is the units digit.

2. According to the problem, $n$ is four times the hundreds digit $h$: $n = 4h$

3. $n$ is also one-third of the "drama digit." Assuming that the "drama digit" is the same as $u$ (the units digit), we get: $n = \frac{u}{3}$

4. The number formed by exchanging the tens and units digits is $100h + 10u + t$. This new number is 18 more than the original number, which gives the equation: $100h + 10u + t = (100h + 10t + u) + 18$

Step-by-step solution:

Step 1: Set up the equation from the exchange of digits.

Simplifying the above: $100h + 10u + t = 100h + 10t + u + 18$ Canceling out $100h$ on both sides: $10u + t = 10t + u + 18$ Simplifying further: $9u - 9t = 18$ $u - t = 2$ Thus, $u = t + 2$.

Step 2: Combine with the condition for $n$.

From the first two conditions: $n = 4h \quad \text{and} \quad n = \frac{u}{3}$ Therefore: $4h = \frac{u}{3}$ Multiplying both sides by 3 to eliminate the fraction: $12h = u$ Since $u = t + 2$, we also have: $12h = t + 2$

Step 3: Solve the system of equations.

Now we have two equations:

1. $u = t + 2$
2. $12h = t + 2$

Let’s substitute $t + 2$ for $u$ into $n = \frac{u}{3}$: $n = \frac{t + 2}{3}$ And since $n = 4h$, we equate this to get: $4h = \frac{t + 2}{3}$ Multiplying through by 3: $12h = t + 2$ This matches the earlier equation, confirming that $12h = t + 2$.

Step 4: Find the value of $n$.

At this point, we can test different values of $h$ to satisfy this condition:

• If $h = 1$, then $12(1) = t + 2$ gives $t = 10$, which is not possible for a digit.
• If $h = 2$, then $12(2) = t + 2$ gives $t = 22$, which is also not possible.
• If $h = 0$, then $12(0) = t + 2$ gives $t = -2$, not possible.

Step 5: No real solution under reasonable assumptions.

It seems there's an inconsistency. Would you like to check further steps? Possible to check