Brain Teasers
Rubber Band Walking
There is a rubber band that is looped around an immovable steel rod. It is three inches long. At the fixed end is a spider. He starts walking along the rubber band at the rate of one inch per minute.
A minute after he started, the rubber band is stretched by three inches. The spider continues walking on it at the same rate.
In another minute, the rubber band is stretched another three inches.
This keeps going on until the spider reaches the end of the rubber band. How long did it take the spider to reach the end?
A minute after he started, the rubber band is stretched by three inches. The spider continues walking on it at the same rate.
In another minute, the rubber band is stretched another three inches.
This keeps going on until the spider reaches the end of the rubber band. How long did it take the spider to reach the end?
Hint
The spider's position is not changed when the rubber band is stretched, but the total distance of travel does change.Answer
10.781355 minutes.The following shows the lengths of the rubber band, the time in minutes, and the distance traveled by the spider.
3, 0, 0
6, 1, 2
9, 2, 4.5
12, 3, 7.333333
15, 4, 10.41667
18, 5, 13.7
21, 6, 17.15
24, 7, 20.74286
27, 8, 24.46071
30, 9, 28.28968
33, 10, 32.21865
36, 10.781355, 36
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Comments
Good teaser - as long as that spider keeps on walking way far away from me!
Yay fred!
also, i hope your teaser about me being smart gets accepted
also, i hope your teaser about me being smart gets accepted
that is a very bored spider very bored
origanial teaser though
origanial teaser though
Hey, I got the answer, but only by using the "long way". Is there a formula I could use to get the answer? Great teaser by the way!
WAY beyond my math capacity!! great teaser!!! u stumped me!
Great teaser! I couldn't get my pet spider to cooperate, so mathematically I simplified the equations down to (representing a move and stretch)
(2/1)a - 1 = b
(3/2)b - 1 = c
(4/3)c - 1 = d
Following that until
(10/9)j - 1 = .7103...
The next move reaching the end before the end of the minute.
Anybody have a simpler way to do it?
(2/1)a - 1 = b
(3/2)b - 1 = c
(4/3)c - 1 = d
Following that until
(10/9)j - 1 = .7103...
The next move reaching the end before the end of the minute.
Anybody have a simpler way to do it?
It took me nearly 10 minutes to figure out its answer nice one though.
good question...but the answer is arguable...since the spider took less than 11 minutes, why did it have to complete a distance of 36 cm? it should be less than that whatever periodic nature the extention took up, rite? if it stretches at 1 minute interval, the spider should reach the other end earlier at a measure of 33 cm, but even if it was a simultanously continuous process as of the spider's walking, the length between terminals still will fall in the 33~36 range (coz the time taken is 10
good question...but the answer is arguable...since the spider took less than 11 minutes, why did it have to complete a distance of 36 cm? it should be less than that whatever periodic nature the extention took up, rite? if it stretches at 1 minute interval, the spider should reach the other end earlier at a measure of 33 cm, but even if it was a simultanously continuous process as of the spider's walking, the length between terminals still will fall in the 33~36 range (coz the time taken is 10
even if it was a simultanously continuous process as of the spider's walking, the length between terminals still will fall in the 33~36 range (coz the time taken is 10
good question...but the answer is arguable...since the spider took less than 11 minutes, why did it have to complete a distance of 36 cm? it should be less than that whatever periodic nature the extention took up, rite? if it stretches at 1 minute interval, the spider should reach the other end earlier at a measure of 33 cm, but even if it was a simultanously continuous process as of the spider's walking, the length between terminals still will fall in the 33~36 range (coz the time taken is 10 < t < 11)...i'd be grateful if you would explain further to clear things up (mb it's just me n my confusion, is all... )
Nice application of series. Didn't quite understand why at first but then it hit me that the rubber band stretches 3 units along its entire length so teh bit infront of teh spider that stretches gets smaller and smaller. Terrific teaser!
Very intriguing, and as Jimbo, took me a second to grasp. Didn't follow it out to its logical end, but figured I could if desired. Nicely done, bot once grasped, not too too difficult.
Great teaser!
Surprising result. The spider's maximum distance from the end of the band occurs after the third stretching when it is 4 2/3 inches away.
Surprising result. The spider's maximum distance from the end of the band occurs after the third stretching when it is 4 2/3 inches away.
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