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# Comments on: FC. Cool Clear Water

## The Math Factor Podcast Site

Last Build Date: Fri, 08 Aug 2014 12:52:06 +0000

By: pamdemonia

Tue, 31 Mar 2009 06:32:32 +0000

I just discovered your podcast, and was working through various problems, and was glad to discover that I got this one right!  my general solution looks a little different, however (brute-forced the 4, 5, and 6 day solutions, then looked for a pattern). It is the summation from 4 to i=n (n being the number of days of the journey) of 3^(i-4). so for n=5 we would have          3^(4-4) + 3^(5-4) =  3^0 + 3^1 = 1 + 3 = 4. Although now that I look again at what y'all were doing above, it's the same thing. I never was very good at converting a summation to a formula.  I did, however, using my formula, come up with the answer to 10 days (1093) in a little over 45 minutes, with no calculator! And then, when listening to the answer on the next podcast, turned on my computer and found this website to point out the mistake!  of course, too late.  but thank you all the same for allowing me to get all excited about math and share it with others who (hopefully) won't give me a (virtual) look of confusion/annoyance I usually get when talking about some math thing to somebody! pam

By: Sean Mc

Mon, 09 Mar 2009 08:10:40 +0000

Oops.  I made the exact opposite error I made the first time.  Now the trekker will have 6 extra units of water stored that never get used since he doesn't head back (one unit of water stored at each of days 1 through 6).  My first answer of 729 incorrectly assumed the trekker used no water himself for the first 6 days.  My second answer of 1457 incorrectly assumed the trekker stored an extra 6 days of water for the return journey.  So its starting to look like your answer of 1093, which is right in between is correct.

By: Sean Mc

Mon, 09 Mar 2009 07:45:14 +0000

Sorry, working without a calculator.  2*(3^6) -1 = 1457

By: Sean Mc

Mon, 09 Mar 2009 07:40:57 +0000

By: strauss

Sun, 08 Mar 2009 06:15:44 +0000

You know, I think you're right. Undoubtedly then the solution we gave is the sum 3^6+3^5+...+1  (Of course it's been a few years, so I don't recall the thinking of the time) [EDIT] No, wait a second, let's think this through. I think you're missing that the porters can't just die of thirst out there but have to have water left along the return route (What kind of a madman ARE you?!!) Before getting started, remember that the geometric series 3^(n-1) + ... + 3 + 1 = (3^n - 1)/2 Suppose we need to go 3 days; that's easy, just go by yourself.
```
3 . . .

. 2 . .

. . 1 .

. . . Victory!
```
4 days isn't too bad with one porter:
```3 . . . .
3 . . . .

. 2 . . .
. 2 . . .

(transferring water:)

. 1 . . .
. 3 . . .

0 . . . .
. . 2 . .   etc

```
Five days starts to get tricky; we need a minimum of four porters:
```3 . . . . .
3 . . . . .
3 . . . . .
3 . . . . .
3 . . . . .

. 2 . . . .
. 2 . . . .
. 2 . . . .
. 2 . . . .
. 2 . . . .

. 1 . . . .
. 1 . . . .
. 2 . . . .
. 3 . . . .
. 3 . . . .

0 . . . . . (these three go back home)
0 . . . . .
0 1 . . . . (but a cache has been left)
. . 2 . . .
. . 2 . . .

0 . . . . .
0 . . . . .
0 1 . . . .
. . 1 . . .  (now this guy can make it back home)
. . 3 . . .  (and this one can head off to the end)
```
What's happening is that the 3^n are supporting the efforts of the all the others. How many porters will it take to go 6? Well, we need to get five people to day 1 with full loads of water. Moreover, we need to have water waiting for the four of those people who will be heading back that way thirsty. So nine more porters will be needed, just to deliver water to that point, for a total of 13 porters and the 1 journeyer. How many people will be needed to go 7 days? Same analysis: we need 14 people to be fully loaded after one day, and a cache of 13 days water ready for the thirteen people who will return. So we need 27 porters just to deliver this extra water and return home themselves. Total of 40 = 27 + 9 + 3 + 1 porters, plus the one person who heads to the end. Etc. Inductively we see that to go N+4 days, we need 1+3+...+3^(N-1) = (3^N - 1)/2 porters. For suppose we need (3^(N-1) - 1)/2 porters to go N+3 days. Then we need (3^(N-1) - 1)/2 + 1 porters to deliver enough water so everyone is fully stocked after one day, and another (3^(N-1) - 1)/2 porters to leave a day's worth of water for the returnees. So we will need an additional (3^(N-1) - 1)/2 + (3^(N-1) - 1)/2 + 1 = 3^(N-1) porters; adding these to the (3^(N-1) - 1)/2 we're already bringing along we have (3^N - 1)/2. qed Thanks!

By: Sean Mc

Sun, 08 Mar 2009 06:01:57 +0000

I'm not sure exactly what number you came up with in your solution to the challenge problem.  I think it was something like 1,096. I get 729, which is 3^6. So at mile 0, you have 729 porters; At mile 1, you have fully-loaded 243 porters and 243 extra days of water; At mile 2 you have 81 fully-loaded porters and 81 extra days of water; At mile 3 you have 27 fully-loaded porters and 27 extra days of water; At mile 4 you have 9 fully-loaded porters and 9 extra days of water; At mile 5 you have 3 fully-loaded porters and 3 extra days of water; At mile 6 you have 1 fully-loaded porter and 1 extra day of water; At mile 7 you have no porters, but can make it 3 miles on your own.