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Problem Statement:

The problem of the week states how many bananas can corey the camel get to the market if he has to eat one banana every mile and it s 1000 miles to the market and he has 3000 bananas and he is able to hold only 1000 bananas at a time.
Process:
I knew that corey had to eat one banana every mile and he had to go 1000 miles but could only carry 1000 bananas at a time and there was 3000 miles so i knew he would have to drop off bananas at certain places to be able to go back and get more bananas. I was writing different amount of miles corey could and how many bananas he had to drop off and pick up and how he could go back and get more bananas from the place where he started. This is the process i did to get my answer.
1. start trip with 1000 bananas
2. travel 200 miles, you're left with 800 - stash 600 at 200 mile point, keep 200 for 200 mile trip back.
3. pick up another 1000
4. travel 200 miles, you have 800 left, pick up 200 from stashed, you now carry 1000 and have 400 more stashed.
5. travel an additional 333 ⅓ miles, you're left with 666 ⅔ , stash 333 ⅓ there (533 ⅓ mile point), you have 333 ⅓ left
6. travel back 333 ⅓ miles to 200 mile point, you have no bananas left, pick up 200 stashed (leaving 200 still at 200 mile point), go back the other 200 miles.
7. pick up another 1000
8. travel to 200 mile point, leaving 800 bananas, pick up remaining 200 stashed
9. with 1000 bananas travel 333 ⅓ miles to 533 ⅓ mile point, you're left with 666 ⅔ bananas.
10. pick up all 333 ⅓ that were stashed there
11. you're back at 1000 bananas
12. make remaining 466 ⅔ mile trip, 1000-466 2/3 = 533 ⅓ bananas left at end.
Solution:
The answer i got was 533 ⅓ bananas will get to the market place in the end. look above for the solution, I started the trip with 1000 bananas the most the camel could carry at once then when i got the 200 mark i dropped off 600 and returned to the beggining and then i picked up another 1000

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