# How many bananas can the camel carry?

This Riddler puzzle is a simple twist on a classic.

You have a camel and 3,000 bananas. You would like to sell your bananas at the bazaar 1,000 miles away. Your loyal camel can carry at most 1,000 bananas at a time. However, it has an insatiable appetite and quite the nose for bananas — if you have bananas with you, it will demand one banana per mile traveled. In the absence of bananas on his back, it will happily walk as far as needed to get more bananas, loyal beast that it is. What should you do to get the largest number of bananas to the bazaar? What is that number?

Here is my solution.
[Show Solution]

## 5 thoughts on “How many bananas can the camel carry?”

1. Dmytro Taranovsky says:

Here is my extension to the puzzle:
The destination is three months away, and we can carry one month of food on the camel. We cannot buy food midway, and unprotected unattended food continuously disappears, with its quantity halving every month. However, we have a magic bag that can fit and protect an unlimited amount of food. Theoretically, can we get to the destination? If yes, about how many months will it take? What if the distance is 6 months?

1. Dmytro Taranovsky says:

It turns out that this is possible but it takes an iterated exponential amount of time, with an extra exponent for each additional 1/2 month. Using a computer to estimate optimal behavior for small distances, I estimated that 3 months distance will take about 10^210000000 months, and for 6 months, we would need about 10^10^10^10^10^10^10^210000000 months of time.

2. I thought it would be more interesting to find the next-to-leading correction to the efficiency which you found. For a starting number (1000 n) and distance (1000 x), we have
n_final = 1000 ( (n+1/2) exp(-x) – 1/2 ) + order(1/n)
I find this by solving the differential equation
dn/dx = -ceiling(n)
using the appoximation ceiling(n) = n+1/2.
With this approximation one would get 787.6 bananas in the original problem, closer to the correct answer 833.3 than the leading-order estimate 3000/e=1103.6.
The order(1/n) correction is much harder to find, as it depends on exactly how far the starting and ending n-values are from integers, and it is not uniform in sign.