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Home » So total odd count = 0 → even → valid - AMAZONAWS

So total odd count = 0 → even → valid - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 5Question: A virtual reality game designer is creating a puzzle where players must align two rotating gears with 24 and 36 teeth respectively. What is the smallest number of full rotations each gear must make before both gears align in their starting position?

📌 Solution: To find when both gears align, we need the least common multiple (LCM) of their number of teeth, 24 and 36. First, factor both numbers:

📌 $24 = 2^3 \cdot 3$,

📌 $36 = 2^2 \cdot 3^2$.

📌 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:

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Question: An ornithologist observes a flock of birds migrating, where...
Solution: The number of birds follows a geometric sequence: $ 3, 6, 12...
The number of birds on day $ n $ is $ 3 \cdot 2^{n-1} $.
\cdot 2^{n-1} > 500
2^{n-1} > \frac{500}{3} \approx 166.67
Now compute powers of 2:
$ 2^7 = 128 $, $ 2^8 = 256 $
So $ 2^{n-1} > 166.67 $ implies $ n-1 \geq 8 $, thus $ n \geq 9 $.
On day 9, the count is $ 3 \cdot 2^8 = 3 \cdot 256 = 768 > 500 $.
On day 8: $ 3 \cdot 2^7 = 384 < 500 $, so day 9 is the first.

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