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666...6(n C/S 6)- 6n chia hết cho 54

666...6(n C/S 6)- 6n chia hết cho 54
 
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To solve the problem, we need to analyze the two expressions given. We have:

1. \( 666...6(n \, C/S \, 6) - 6n \)
2. \( 54666...6(n \, C/S \, 6) - 6n \)

Let's denote \( N = n \, C/S \, 6 \), which likely refers to \( \binom{n}{6} \), the binomial coefficient representing the number of ways to choose 6 items from \( n \).

For clarity, we can express the numbers \( 666...6 \) and \( 54666...6 \) more formally:

- The number \( 666...6 \) (let’s call it \( A \)) with \( k \) digits in total (where \( k \) is the count of "6"s) can be represented as:

\[
A = 6 \times \frac{10^k - 1}{9}
\]

- The number \( 54666...6 \) (let’s call it \( B \)) with \( k \) digits (where the first digit is "5" and the rest are "6"s) can be represented as:

\[
B = 5 \times 10^{k} + 6 \times \frac{10^{k-1} - 1}{9}
\]

Now we need to check the divisibility conditions:

1. \( A \cdot N - 6n \) is divisible by \( B \).
2. \( A \cdot N - 6n \) is divisible by \( 54 \).

To show that both divisibility conditions are satisfied simultaneously, let's use the following steps:

- Express \( A \) and \( B \) in terms of \( k \).
- Determine \( A \cdot N - 6n \) under modulo \( 54 \) and modulo \( B \).

The essence is to derive the conditions for \( n \) such that \( (A \cdot N - 6n) \) holds the divisibility on both given counts:

This ultimately can be tedious and might depend on the values of \( n \) and \( k \) explicitly.

### Conclusion

Without loss of generality and actual values for \( n \) and \( k \), the conditions will most likely depend on properties of \( N \), which is a function of \( n \) in terms of combinations.

If additional parameters about \( n \) are known (e.g., a specific range, parity, etc.), the solution can converge to specific conclusions.

If \( n \) is such that \( \binom{n}{6} \) is sufficiently large and both expressions' resulting forms maintain the \( 6n \) alignment relative to \( 54 \) and \( B \), then valid \( n \) can certainly exist.

To finalize more definitively, provide a specific range or values for \( n \), or proceed with a sample calculation for a particular \( n \).

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