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Topic: Bus passengers (Read 1012 times) |
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benignuman
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Bus passengers
« on: May 11th, 2008, 8:02pm » |
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I am having a tremendous amount of trouble with this problem
" The UC Berkeley bus had a minimal number of passengers. When it arrived at Telegraph Avenue, 3/4 of the passengers got out, and 7 people got on. At the next two stops, Shattuck and Hearst, the same thing happened. How many got off at Hearst"
It seems to me to be impossible.
Au secours!
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william wu
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Re: Bus passengers
« Reply #1 on: May 11th, 2008, 8:45pm » |
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Here's how I did it; perhaps there is a better way.
Let m_k be the number of passengers on the bus after the kth stop. Let r_k be the number of passengers that leave at the kth stop. Then we have the equations
m_{k+1} = (1/4)(m_k) + 7
r_{k+1} = (3/4)(m_k)
We let m denote the number of passengers initially on the bus. We wish to solve for r_3.
Writing out m_1,m_2,m_3,r_1,r_2, and r_3 all in terms of m yields
m_1 = 7 + m/4
m_2 = (140+m)/16
m_3 = (588+m)/64
r_1 = 3m/4
r_2 = 3(28 + m)/16
r_3 = 3(140+m)/64
where all of these quantities must be integers. Examining the m_3 equation, we see that 588 + m must be a multiple of 64. The smallest positive integer we can add to 588 to satisfy this constraint is m=52. Plugging m=52 into the other equations results in integer quantities as well, so we have found the minimal value for m. Lastly, r_3 = 3(140+52)/64 = 9.
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Re: Bus passengers
« Reply #2 on: May 12th, 2008, 3:25am » |
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((m/4+7)/4+7)*3/4 =
3m/64 + 21/16 + 21/4 =
(3m+420)/64
Because it has to be integer, 3m+420 must be divisible by 64,
3m+420 = 3m + 36 = 0 (mod 64), therefore m = -12 (mod 64)
minimum positive m is -12+64=52
And we get (3*52+420)/64 = 9
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Wikipedia, Google, Mathworld, Integer sequence DB
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benignuman
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Re: Bus passengers
« Reply #3 on: May 12th, 2008, 4:51am » |
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Thanks people. I was thrown off by the "minimal passengers" at the beginning I assumed it had to start with a small amount of people.
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rmsgrey
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Re: Bus passengers
« Reply #4 on: May 12th, 2008, 5:10am » |
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on May 12th, 2008, 4:51am, benignuman wrote:| Thanks people. I was thrown off by the "minimal passengers" at the beginning I assumed it had to start with a small amount of people. |
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As you've probably figured out, it's not that the bus started with a small number of passengers, but that it started with the smallest number that made the whole process possible.
Given any initial number of passengers, m, for which the process works, m+64 will also work, so there are an infinite number of initial passenger loads for which the process would work (though, depending on the bus, even the 116 passengers of the next solution up from the minimal may have trouble fitting on - London double-decker buses used to be rated for 144 passengers maximum - this was before they got redesigned with fewer seats, which reduced the carrying capacity)
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Sealock
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Re: Bus passengers
« Reply #5 on: May 12th, 2008, 8:01am » |
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And what is the name of the bus driver?
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