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Topic: Workmates' Lunchtime (Read 1638 times) |
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ThudnBlunder
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Workmates' Lunchtime
« on: Apr 8th, 2005, 2:49am » |
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Two workmates each take one lunch break per day. They arrive at the restaurant independently at (uniformly) random times between 12pm and 1pm, and each stays for exactly x minutes. The probability that either workmate arrives while the other is in the restaurant is 2/5.
If x = a - bc1/2 find the value of a + b + c
where a,b,c are positive integers and c is indivisible by the square of any prime.
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| « Last Edit: Apr 8th, 2005, 8:28am by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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Sjoerd Job Postmus
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Re: Workmates' Lunchtime
« Reply #1 on: Apr 8th, 2005, 3:31am » |
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A start:
awowip:Arrives while other workmate is present
PA_awowip = 1 / 5
PB_awowip = 1 / 5
From this follows that they both stay exactly 12 minutes. ( A enters while B present + B enters while A present )
Now, to fill in the formula... 
Modification: Not divisable by the square of any prime. So c != dp2, right? but c might be the square of a prime... (same goes for a and b, but I'm more concerned about c)
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| « Last Edit: Apr 8th, 2005, 3:33am by Sjoerd Job Postmus » |
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Grimbal
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Re: Workmates' Lunchtime
« Reply #2 on: Apr 8th, 2005, 9:04am » |
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1. | hidden: |
If x and y are the arrival time in hours, then the "collisions" cases look like this:
...../####
.../#####/
./######/.
/#####/...
####/.....
You can join the 2 empty areas to get:
##########
##########
####......
####......
####......
So the collision surface is 12-(1-x)2
Solving 12-(1-x)2 = 2/5 gives
x2 - 2x + 2/5 = 0
or
x = (2 - [sqrt](4-8/5))/2
x = 1 - sqrt(3/5) = 0.225566 hours = 13 min 32 s
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ThudnBlunder
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Re: Workmates' Lunchtime
« Reply #3 on: Apr 8th, 2005, 9:07am » |
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on Apr 8th, 2005, 3:31am, Sjoerd Job Postmus wrote:A start:
From this follows that they both stay exactly 12 minutes.
Modification: Not divisable by the square of any prime. So c != dp2, right? but c might be the square of a prime. |
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PSB (and Grimbal?), it is not stipulated that they must leave before 1pm.
If c is a square of a prime then it is divisible by itself, a square of a prime. 
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| « Last Edit: Apr 9th, 2005, 7:53am by ThudnBlunder » |
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Grimbal
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Re: Workmates' Lunchtime
« Reply #4 on: Apr 8th, 2005, 9:17am » |
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If they had to leave before 1pm, they couldn't stay x minutes and arrive uniformly between 12 and 1.
I only consider arrival times. If the difference between their arrival times is less than x, then they meet..
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Grimbal
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Re: Workmates' Lunchtime
« Reply #5 on: Apr 8th, 2005, 9:22am » |
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AAAh. Now I understand what was asked.
x = 1 - sqrt(3/5) hours. = 60 - 12*sqrt(15) minutes
a+b+c = 87
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| « Last Edit: Apr 8th, 2005, 9:28am by Grimbal » |
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Eigenray
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Re: Workmates' Lunchtime
« Reply #6 on: Apr 8th, 2005, 12:25pm » |
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(If you want n coworkers to all meet, see Clustered Points.)
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