"Carrie the Camel works for the owner of a small, remote banana plantation. This year's harvest consists of three thousand bananas. Carrie can carry up to one thousand bananas at a time. The market place where the bananas are sold is one thousand miles away. Unfortunately, Carrie eats one banana at the beginning of every mile that she walks (when she is carrying the bananas). Of the three thousand bananas harvested, what is the largest number of bananas Carrie can get to market? Hint: Assume that any pile of bananas Carrie happens to leave off somewhere between the plantations and the market won't be tampered with while she is away; when she returns to the pile, the same number of bananas that she left off will be waiting for her.

Solution to the Problem: The most bananas that Carrie can get to market appears to be 833. When I first posted this problem, I believed that Carrie could get at most 750 bananas to the market. However, several of you sent in the solution of 833, and it worked! I was afraid that when I retired, my brain would turn to mush, and maybe that's what happened!! Anyway, here is James Alarie's excellent explanation: Carrying any number of bananas up to 1000 any distance forward costs 1 banana per mile. No matter how you do it, moving 1001-2000 bananas forward costs 2 bananas per mile, and 2001-3000 bananas costs 3 bananas per mile. You lose 999 bananas in the first 333 miles, 1000 bananas in the next 500 miles, and 167 bananas in the remainder of the trip. This leaves you with 834 bananas. To actually carry out the above move: 1. pick up 1000 bananas from the plantation, move them 333 miles toward the market, and leave the remaining 667 bananas at that point. 2. go back to the plantation, pick up another 1000 bananas, move them 333 miles toward the market, find the 667 bananas that you left there on the previous trip, pick up 333 of them so that you now have 1000 bananas on Carrie, move forward another 500 miles, and drop the remaining 500 bananas there. 3. go back to the plantation, pick up 999 bananas, move them 333 miles toward the market, find the 334 left there on the previous trip, add them to the load of 666 bananas already on Carrie for a total of 1000, move forward 500 miles, pick up the 500 bananas left there on the previous trip so that you now have 1000 bananas on Carrie, travel the remaining 167 miles to market, deliver 833 bananas to market. 4. go back to the plantation and eat the remaining banana yourself. James Alarie took the problem to another level. He asked, "What if you have 4000 bananas available? Or 5000, 6000, ...? Is there an equation for this?" He then sent in the following analysis: I based my work on the idea that moving X bananas one more mile toward market will cost (X/1000) bananas per mile rounded up. If I have 3000 bananas at the beginning, then the first mile will cost me three bananas, and each suceeding mile will cost three bananas until the load is down to 2999. Working in that way, I wrote a Javascript function to do the work and called it with ever-increasing numbers of bananas to see what would happen. To my surprise, not only did the number of bananas delivered to market go over 1/3 of what I started with, but the percentage maxed out and then started dropping at 52000 bananas! The best delivery ratio which I could obtain was 35.698% with the following funciton: function Market(K) { M=0;' // mile marker while ((M+(D=Math.floor(1000/D))) < 1000) { M=M+D; K=K-1; } return M*K;

Correctly solved by:

The following persons sent in answers of 750 -- the same number that I originally had: