Math Puzzles (Read the Rules!)

[ChUcK]

I don't have time to really work on it, but I can get 500 lbs of grain to the destination with a lot left over at the start and a pissed-off, stranded camel. I can't yet figure out how to get the camel to the destination twice.

[Timko]

That's more than I can get there. You need to take smaller "steps." So move the grain all the grain to spot 1, then all the grain to spot 2, etc.

[Parks]

Going off of Chiggins hint that max efficiency is carrying 1,000 lbs at a time, I can get about 1,400 lbs of grain to the city.

Basically, you need to move 10,000 lbs at the start, and the most efficient way to do this based on the hint would be 10 trips to the cache. This would be 19 half trips, since you don't have to return to the start once you've taken the last load to the cache.

Since carrying 1,000 lbs at a time is max efficiency, you want your cache sites to have a number of lbs evenly divisible by that amount. For reasons I'll explain in a second, your cache sizes should each be 1,000 lbs less than the last, so the first cache will be 9,000 lbs as Chiggins explained.

So your first cache trip looks like: (10,000lbs x 19z1 mile trips) = 9,000lbs, so z1 = ~52.6 miles to the first cache.

Then your next trip you want to move 9,000lbs to an 8,000lbs cache z2 miles away. Now, to move 9,000lbs, you need 17 trips (2 x (grain at start / 1000 lbs) - 1). So the number of trips is decreasing at each step. This is why you want to only cache in 1,000lb increments, because your efficiency goes up as your cache size decreases. So, (9,000lbs x 17z2 mile trips) = 8,000lbs leaves us with z2 at 58.8 miles to the 2nd cache from the first, for a total of ~111.4mi traveled.

Continue in this manner until we are at about 800 miles (rounding off, I came to 799.8m) with 2,000lbs in your final cache. Then the city is within reach so you can make a two way trip at a cost of 400lbs of grain, leaving 600 lbs at the city and 1,000 in your last cache. Then you make a one way trip for 200 lbs and you get another 800lbs to the city, for a total of 1,400lbs of grain.

I'm certain there's a much more elegant way to represent the math, but this was just kind of following my thought process without formalizing it.

I'm pretty sure this is correct, but there could be a better way so I won't post my puzzle until I get confirmation on this. [edit]Searching online, it looks like this is right, so I'll post my puzzle.[/edit]

[chiggins]

Going off of Chiggins hint that max efficiency is carrying 1,000 lbs at a time, I can get about 1,400 lbs of grain to the city.

Ed McMahon says: You are correct, sir!



Nice work, I'd be beating my head against the wall until Christmas of next year trying to crack that one.

I'm certain there's a much more elegant way to represent the math, but this was just kind of following my thought process without formalizing it.

I'm pretty sure this is correct, but there could be a better way so I won't post my puzzle until I get confirmation on this.

Nope, that is exactly how it's represented. Here's the solution, and it's formatted about the way you did it.

Your move!

[Parks]

You have a lighter and two ropes which each take 60 minutes to burn from one end to the other. However, the ropes do not burn at a uniform rate, so it might take 50 minutes to burn 10% of the rope, and then 10 minutes to burn the other 90%. Also, the first rope burns at different rates from the second rope, so you can't compare the two by looking at their relative lengths.

Using only these three items, how can you measure exactly 45 minutes?

[Luke]

Even though I'm banned, I'll answer this one.

Light one end of one rope, and both ends of the second rope.

Then, when the second rope is all gone, at 30 minutes, light the the other end of the first rope, which, going double time will only have 15 minutes left.


Here is an easy one, but I don't know too many off hand.

You go to the store, averaging 60 miles an hour. On the way back, you take the exact same route, except there is traffic and you average 40 miles an hour. What is your average speed during the trip?

[jubuttib]

Buh, hope anyone didn't see my last post before I deleted it. Anyway, I think it's 48 mph, but I don't want to come up with the next one, at least tonight, so I'll leave my reasoning out.

Not counting the time spent in the store of course, since it wasn't specified, just the traveling part of the trip.

[ChUcK]

48 is indeed the correct answer.

Assume a 20 mile round trip.

First 10 miles are covered averaging 60mph => takes 10 minutes
Second 10 miles covered averaging 40mph => takes 15 minutes.

20mi/25min = 48 mph.

[ChUcK]

Bukowski is thinking of hiring you to help him bet on the ponies at the track. In order to get the job (and who wouldn't want to be his assistant) you have to solve this problem:

There are 16 horses, each with different constant speeds. The quantitative speeds are unknown, but qualitatively the horses never change speeds. With only your wits, and a racetrack that can run heats of four horses maximum at once, figure out the four fastest horses. No stopwatches are in sight.

You'll only get the job if you can find the horses quickly (fewest possible # of heats) because Bukowski is currently sober and if you take too long to figure it out, he'll head to the bar without you and hire the first bum he sees. This is your big chance!

[Luke]

That's a good one, and I think I got it, but I want to see some other thoughts on it and I don't want to think of another puzzle.

[ChUcK]

PM me your answer, I always enjoy seeing the way this problem is solved by various personalities.

[Timko]

Damn, I'm totally falling asleep on my own thread. I'm having too much fun reading the lolz in Up In Smoke.

[veganray]

Run four random horses in heat one. Top 3 run again & fourth is put into a holding pen.
Add one of the yet-to-run horses for heat two. Top 3 run again & fourth is put into the holding pen.
Repeat this process until every horse has run at least once (13 heats).
This will yield the three fastest horses, finishers 1, 2, & 3 in heat thirteen.

To find the 4th fastest, run four random horses from the holding pen in heat one. This time, only the winner runs again & the 2nd, 3rd, & 4th place horses are summarily euthanized.
Repeat this second process, removing three horses from the holding pen each time) until only one horse (the winner of the fourth & final heat) is left alive. This is the 4th fastest horse overall.

This method will yield the fastest 4 horses in 17 heats.

[veganray]

I'll bet you $100 that if you give me $200, I will give you $300 in return.

Would you bet with me? Why?

[chiggins]

With only your wits, and a racetrack that can run heats of four horses maximum at once, figure out the four fastest horses.

Question: are we figuring out which four horses are the fastest, and the order of those four, or are we simply finding which four are the fastest without regard to how those four finish?