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Home » So no integer \( n \) gives 210? But the problem must have a solution. - jntua results

So no integer \( n \) gives 210? But the problem must have a solution. - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 So the structure is uniquely determined: H in positions 1,3,5,7 and P in 2,4,6.

📌 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 positions (with 3 P’s in between).

📌 This is the only possible arrangement (up to shifting, but shifting is fixed by positions).

📌 So only one valid configuration exists.

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We are given the quadratic expression:
4x^2 - 12x + 9
Check if it is a perfect square trinomial. Identify:
First term: \( (2x)^2 = 4x^2 \)
Last term: \( 9 = 3^2 \)
Middle term: \( -12x = -2 \cdot 2x \cdot 3 \)
4x^2 - 12x + 9 = (2x - 3)^2
\boxed{(2x - 3)^2}
Two proposed transit routes are modeled by the lines \( 3x - 4y = 12 \...
3x - 4y = 12 \quad \text{(1)}\\

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