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Home » In Print We Trust: These 10 Brands Prove Why Physical Media Still Wins! - Londonproperty

In Print We Trust: These 10 Brands Prove Why Physical Media Still Wins! - Londonproperty

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 $36 = 2^2 \cdot 3^2$.

📌 The LCM is found by taking the highest powers of all primes:

📌 $ ext{LCM} = 2^3 \cdot 3^2 = 8 \cdot 9 = 72$.

📌 Thus, the smallest number of teeth rotations needed for alignment is 72. For the first gear (24 teeth), this corresponds to $72 \div 24 = 3$ rotations, and for the second gear, $72 \div 36 = 2$ rotations.

📌 Therefore, the smallest number of full rotations for alignment is $oxed{3}$ rotations for the 24-tooth gear and $oxed{2}$ for the 36-tooth gear, but the alignment occurs at $oxed{72}$ total teeth moved, so the minimal number of full rotations of the *first gear* required is $oxed{3}$. However, the question asks for the smallest number of full rotations each must make to alignâthis is interpreted as the LCM of their rotation cycles. Since one full rotation of the 24-tooth gear moves 24 teeth, and 72 is the LCM, the first gear rotates $72/24 = oxed{3}$ times, the second $72/36 = oxed{2}$. But the smallest number of full rotations each must make to realign is the LCM cycle completed, so the answer is the LCM of the number of rotations: $3$ rotations for the 24-tooth gear and $2$ for the other. But to match the formatâsingle answerâwe interpret as:

📌 The smallest number of full rotations of the **24-tooth gear** required for alignment (so that both complete an integer number of cycles) is $oxed{3}$.

📌 Thus, the final answer is $oxed{3}$.

📌 Question: A soil scientist is analyzing soil samples from three different fields, each sampled every 6, 8, and 12 days respectively. After how many days will all three fields be sampled on the same day again for the first time?

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So after 14 hours, it drops below 100,000.
Each 2 hours halves → decay factor = 0.5 every 2 hours
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Take log: n × log(0.5) < log(0.0125)
n > log(0.0125)/log(0.5) ≈ (-1.903)/(-0.3010) ≈ 6.32 → n ≥ 7 halvings
× 2 = 14 hours
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