# IMP: Year 1: POW J

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• Oct 16th 2007, 04:11 PM
JourneyingTogether
IMP: Year 1: POW J
I have just begun the IMP program for the school year for 8th graders. All of my friends and I a stuck on this IMP problem. It is one of our POW's or "Problem of the Week" We have started POW J: A Camel Messenger.
It states:
"Dory is a fictitious camel who lives on bananas. She one banana every mile she walks. She can carry at the most 1000 bananas at a time. Dory is being sent to deliver a message. The trip will take her 1000 miles across the desert. She needs to get there, deliver her message, and then return home. She knows that there are no bananas to be found along the route in either direction, but she can also know that she can leave banans safely at any place along her route."
Then it asks:
"What is the minimum number of bananas that Dory needs to have at the start of her journey to reach her destination and return only using those bananas?"
Our teacher says that we can figure it out using the formula. She says that the formula contain one or more of the following tools to solve it: sigmas, square roots, intigers, fractions, dectimals, factorials, logs (half of us don't even know what logs are) and the order of operations (PEMDAS)
Could you help us find a formula?
J.T.
• Oct 24th 2007, 10:11 AM
Mark@Work
Just to get started
Dory travels 250 miles (eating 250 bananas) drops off 500 and walks back (eating the remaining 250)
2nd journey
Travels 250 miles eats 250 bananas, picks up 250 from the stash (250 left in stash 1000 on Dory).
250 more miles drops off 500, (250 left). Returns 250 miles to first stash picks up 250 bananas, enough to get home.
3rd journey same as 1st to re-stock 250 mile stash
4th journey travel 250, re-stock (250 left in 250 mile stash) travel to 500, re-stock (250 left in 500 mile stash, 1000 on Dory), travel to 1000 eating 500, return to 500 eating 500, return home reducing stashes to zero.

I'll leave it for someone else to put this into math terms (sum of a series?) and to prove (or disprove) that this is the most efficient way of stashing.

Mark.
• Oct 25th 2007, 01:51 AM
Mark@Work
An Alternative
Dory travels 250 miles dropping off 3 bananas each mile, and eating 1 banana each mile, return eating one stashed banana each mile.
2 bananas each mile out to 250 miles.
Travel to 250 miles eating 250 bananas, travel from 250 to 500 dropping 2 bananas and eating 1 every mile, return eating 1 stashed banana each mile.
1 banana each mile out to 500 miles.
Repeat these 2 journeys.
2 bananas each mile out to 500 miles.
Travel to 500 miles eating one stashed banana each mile, continue to 1000 miles eating one carried banana each mile, get to destination with 500 bananas. Return to 500 eating these bananas, return the rest of the way eating the last of the stashed bananas.
5 journeys at 1000 bananas per journey.
If I've got this correct my previous solution used 4 journeys, making it more efficient...