n is four times the three digit hundreds digit and one third of the drama digit. If the number exchanged between units and tens digits is 18 more than the previous digit, what is the value of n

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

n is four times the three digit hundreds digit and one third of the drama digit. If the number exchanged between units and tens digits is 18 more than the previous digit, what is the value of n?

This problem involves solving for n in a three-digit number, where n is four times the hundreds digit and one-third of the units digit. It also examines digit exchanges between the tens and units place, providing a detailed breakdown of algebraic equations. Suitable for students in Grades 9-12.

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