To solve this problem, let the number $n$ be expressed in both base 6 and base 9, as $abc$ and $cba$, respectively. We aim to determine the decimal representation of $n$.

Step 1: Represent $n$ in base 6

In base 6, the number $abc$ represents: $n = a \cdot 6^2 + b \cdot 6^1 + c \cdot 6^0$ $n = 36a + 6b + c$

Step 2: Represent $n$ in base 9

In base 9, the number $cba$ represents: $n = c \cdot 9^2 + b \cdot 9^1 + a \cdot 9^0$ $n = 81c + 9b + a$

Step 3: Equate the two representations

Since $n$ is the same in both cases, equate the expressions: $36a + 6b + c = 81c + 9b + a$

Step 4: Rearrange the equation

Simplify the equation: $36a - a + 6b - 9b + c - 81c = 0$ $35a - 3b - 80c = 0$

Step 5: Solve for integer solutions

The digits $a$, $b$, and $c$ must satisfy $0 \leq a, b, c \leq 5$ (since they are base-6 digits). Let us solve for possible values.

Rearrange the equation:

$35a = 3b + 80c$ $a = \frac{3b + 80c}{35}$

For $a$ to be an integer, $3b + 80c$ must be divisible by 35. Testing possible values for $c$:

• Case $c = 1$: $3b + 80(1) = 35a$ $3b + 80 = 35a$ $3b = 35a - 80$ Test values of $a$ such that $3b$ is divisible by 3:
  • $a = 2$: $35(2) - 80 = 70 - 80 = 10$. Not valid. Let's refine this calculation to systematically explore solutions where $a, b, c$ are valid digits (0 to 5).

Step-by-Step Solution:

The key equation is: $35a = 3b + 80c$

Rearrange:

$a = \frac{3b + 80c}{35}$

Since $a$ must be an integer (0 to 5), $3b + 80c$ must be divisible by 35. Let us test feasible values of $c$ (0 to 5) and solve for $b$ and $a$.

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Case $c = 2$:

Substitute $c = 2$: $3b + 80(2) = 3b + 160 = 35a$ $3b = 35a - 160$

Solve for divisibility:

For $a = 5$: $3b = 35(5) - 160 = 175 - 160 = 15$ $b = \frac{15}{3} = 5$

Thus, $a = 5$, $b = 5$, $c = 2$.

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Verify Solution:

For $a = 5, b = 5, c = 2$:

• In base 6: $n = 36a + 6b + c = 36(5) + 6(5) + 2 = 180 + 30 + 2 = 212$.
• In base 9: $n = 81c + 9b + a = 81(2) + 9(5) + 5 = 162 + 45 + 5 = 212$.

The solution is consistent.

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Final Answer:

The decimal representation of $n$ is: $\boxed{212}$

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Would you like further clarification or more exploration? Here are 5 related questions to deepen your understanding:

1. How do positional number systems (like base 6 or base 9) work in general?
2. What are other scenarios where a number appears the same in two bases?
3. How do we solve similar problems involving bases higher than 10?
4. What mathematical properties ensure $a, b, c$ are integers in these equations?
5. How can divisibility rules help in problems involving modular arithmetic?

Tip: Always verify solutions by substituting back into the original base equations!