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 June 15th, 2015, 01:48 AM #1 Senior Member   Joined: Apr 2008 Posts: 193 Thanks: 3 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 03:29 AM.
 June 15th, 2015, 03:44 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 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). Thanks from Country Boy
 June 15th, 2015, 06:30 AM #3 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,134 Thanks: 720 Math Focus: Physics, mathematical modelling, numerical and computational solutions 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 $\displaystyle \frac{24}{8} = 3$ Thanks from Country Boy

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