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Home » The square of \( (3x - 2) \) is \(\boxed9x^2 - 12x + 4\).**Question: - AMAZONAWS

The square of \( (3x - 2) \) is \(\boxed9x^2 - 12x + 4\).**Question: - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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No other configuration allows 4 non-adjacent H’s in 7 positions.
Thus, only **one** such sequence satisfies the condition.
But wait — is that correct? Let's verify:
Can we place 4 non-adjacent H’s in 7 positions?
The minimal length needed for 4 non-adjacent H’s is: H _ H _ H _ H → 7...
This is the only possible arrangement (up to shifting, but shifting is...
So only one valid configuration exists.
But let's use the standard combinatorial method:
To place \(k = 4\) non-adjacent H’s in \(n = 7\) positions:
The number of ways is \(\binom{n - k + 1}{k} = \binom{7 - 4 + 1}{4} =...

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This means \(n^3 \equiv 888 \pmod{8}\) and \(n^3 \equiv 888 \pmod{125}...
Step 1: Solve modulo 8
\(888 \equiv 0 \pmod{8}\), so:
n^3 \equiv 0 \pmod{8} \Rightarrow n \equiv 0 \pmod{2}
In fact, since \(8 = 2^3\), for \(n^3 \equiv 0 \pmod{8}\), \(n\) must...
If \(n \equiv 0 \pmod{2}\), then \(n^3 \equiv 0 \pmod{8}\) only if \(n...
\(2^3 = 8 \equiv 0\)
\(4^3 = 64 \equiv 0\)
\(6^3 = 216 \equiv 0\)
So all even \(n\) satisfy \(n^3 \equiv 0 \pmod{8}\). So condition is \...

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