two gears work together. There are 32 teeth on one gear and 36 on the other gear. The first gear makes 128 revolutions per second. How often, during a 24 hour period, will the same two teeth from each gear be together?
+128475 Two gears work together. There are 32 teeth on one gear and 36 on the other gear. The first gear makes 128 revolutions per second. How often, during a 24 hour period, will the same two teeth from each gear be together?
I saw geno do something like this one the other day......
32 = 2^5
36 = 3^2 * 2^2
The least common multiple between 36 and 32 = 3^2 * 2^5 = 288
So....... the two teeth are together when the first gear makes 288/32 = 9 revolutions and the second gear makes 288/36 = 8 revolutions
And it takes the first gear 9/128 seconds to make 9 revolutions
And there are 60*60*24 = 86400 seconds in a 24 hour period.......so......
In 24 hours.......the same two teeth from both gears will be together
86400 / [ 9/128] = 1,228,800 times
+33616 lcm(32, 36) = 288
revolutions in 24 hours = 128*3600*24
number of times same teeth together = 128*3600*24/288 → 38400
.
+118609
Thanks Alan,
That one had me stumped. I was looking for some really complicated solution :)
+128475 Two gears work together. There are 32 teeth on one gear and 36 on the other gear. The first gear makes 128 revolutions per second. How often, during a 24 hour period, will the same two teeth from each gear be together?
I saw geno do something like this one the other day......
32 = 2^5
36 = 3^2 * 2^2
The least common multiple between 36 and 32 = 3^2 * 2^5 = 288
So....... the two teeth are together when the first gear makes 288/32 = 9 revolutions and the second gear makes 288/36 = 8 revolutions
And it takes the first gear 9/128 seconds to make 9 revolutions
And there are 60*60*24 = 86400 seconds in a 24 hour period.......so......
In 24 hours.......the same two teeth from both gears will be together
86400 / [ 9/128] = 1,228,800 times
