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Home » Since \( t > 0 \), we take the positive root: - Simpleprint

Since \( t > 0 \), we take the positive root: - Simpleprint

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 Since \( z \) is divisible by 7, 11, and 13, compute the least common multiple:

📌 ext{lcm}(7, 11, 13) = 7 imes 11 imes 13 = 1001

📌 But 1001 is a four-digit number. The condition specifies a three-digit number, and no three-digit number divisible by all three exists. However, reconsider: the smallest three-digit number divisible by \( 7 imes 11 imes 13 = 1001 \) is impossible.

📌 But note: the problem asks for the smallest three-digit number divisible by **each** of 7, 11, and 13. Since their product is 1001 > 999, no three-digit number satisfies this.

📌 But perhaps a misinterpretation: maybe the number is divisible by **at least one**, but the context implies "divisible by" the product. Given the constraints, no such three-digit number exists. But rechecking, the smallest number divisible by 7, 11, and 13 is 1001âexceeding three digits.

📌 Waitâperhaps the problem intends the smallest three-digit number **divisible by the lcm**, but since lcm is 1001, no such number exists.

📌 But to resolve: reconsider if "divisible by 7, 11, and 13" means divisible by their **product**âwhich is 1001âso no three-digit multiple.

📌 But if the intended meaning is divisible by **each** individually, then the lcm is still 1001.

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\end{vmatrix} = \mathbf{i}(y \cdot 3 - z \cdot 2) - \mathbf{j}(x \cdot...
= egin{pmatrix} 3y - 2z \ - (3x - z) \ 2x - y \end{pmatrix} = egin{p...
Set equal to \(egin{pmatrix} 0 \ 0 \ 5 \end{pmatrix}\):
3y - 2z = 0 \
-3x + z = 0 \
2x - y = 5
From second equation: \(z = 3x\)
From first: \(3y = 2z = 2(3x) = 6x \Rightarrow y = 2x\)
Substitute into third:
2x - (2x) = 5 \Rightarrow 0 = 5

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