By Alex Cohen

Problem Statement: Corey Camel has 3,000 bananas to take to the market. He can only carry 1,000 at a time. The market is 1,000 miles away. Every mile Corey eats a banana. My job is to find out; how many bananas he can get to the market place?

Process: To do this POW I started with Mini Camel to find a good strategy. The strategy we found was this: take as many bananas as you can to a certain distance and drop the rest, minus enough to get back. For example go 100 miles, drop 800, then come back 100 miles, you have used all of your bananas. Next I applied this method to Corey Camel, and since our teacher said the most number of bananas he can bring back was 533 1/3 I realized that after the 1st 1,000 were gone, I would use the other 2,000 by dividing them into 1/3s since the only way to get 33 1/3 is to divide something by 3, like 1,000 divided by 3 is 333 1/3. 533 333 is 200, so I know that the 1st trip has to evolve 200, since I need 200 to be left over when I'm done to get me the rest of the way. I drew out a time line of events to show Corey's movements as he progresses.

Solution: Corey starts by picking up 1,000 bananas, and walking them 200 miles. Next he drops 600 and goes back 200. After that he picks up another 1,000 and goes 200 miles. He then picks up 200 bananas from his pile of 600. He now has 1,000, which he takes 333 1/3 miles farther to the 533 1/3 market, and drops 333 1/3 and goes back. He picks up 200 bananas and goes back to the start to get the last 1,000. Next he travels 200 back to the last set of 200 on the ground, and picks them up. He is now left with 1,000 bananas on him, with 333 1/3 waiting for him. He goes forward 333 1/3 and then picks them up. He then goes the last 333 1/3 and gets to the market with 533 1/3 bananas. Evaluation: One thing I didn't like about this POW is that with cut out pieces of paper you can solve it, but logic parts of it were helpful. I would say that on a scale...