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June 15th, 2015, 12:48 AM   #1
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Is this word problem solvable with a lack of information?

Two gears engage with each other. Gear 1 has 6 teeth. Gear 2 has 8 teeth. As Gear 1 turns, it causes Gear 2 to turn at a different rate. Gear 1 is rotated until the two gears are back to the starting position. What is the minimum number of rotations Gear 1 requires to return to this starting position?

I have no idea of how to do it. Could someone please explain it? Thanks.

Last edited by skipjack; June 15th, 2015 at 02:29 AM.
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June 15th, 2015, 02:44 AM   #2
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You need to assume the teeth of gear 1 engage directly with those of gear 2, so that each revolution of gear 1 causes 3/4 of a revolution of gear 2. The gears then first return to their starting positions after 4 revolutions of gear 1 (which have caused 3 revolutions of gear 2).
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June 15th, 2015, 05:30 AM   #3
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This is a lowest common multiple question.

prime factor decomposition:

$\displaystyle 6 = 2 \times 3$
$\displaystyle 8 = 2 \times 2 \times 2$

Highest common factor = 2

Therefore, the lowest common multiple is $\displaystyle (2 \times 3) \times (2 \times 2 \times \cancel{2}) = 2 \times 2 \times 2 \times 3 = 24$

So, 24 teeth will connect in order for the gears to return to their starting position.

number of rotations for gear 1 is $\displaystyle \frac{24}{6} = 4$
number of rotations for gear 2 is $\displaystyle \frac{24}{8} = 3$
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