## MATHEMATICA!

# IBM Research: Mathematica Challenge!

by IBM:

You are cordially invited to match wits with some of the best minds in IBM Research.

Seems some of us can’t see a problem without wanting to take a crack at solving it. Does that sound like you? Good. Forge ahead and ponder this month’s problem. We’ll post a new one every month, and allow two to three weeks for you to submit solutions (we may even publish submitted answers, especially if they’re correct). We won’t reply individually to submitted solutions but every few days we will update a list of people who answered correctly. Towards the end of the month, we’ll post the answer.

So give your mind a break from its routine—you never know what other problems you may solve in the process!

June’s Challenge:

This challenge is based on a puzzle we heard from Yaniv Shmueli, who heard it himself a few years ago.

Let’s assume that cars have a length of two units and that they are parked along the circumference of a circle whose length is 100 units, which is marked as 100 segments, each one exactly one unit long.

A car can park on any two adjacent free segments (i.e., it does not need any extra maneuvering space).

Our question is as follows: Let’s assume that we start with an empty circle. We add one car at a time, and each car parks in a random free space (aligned to a unit length), till no such place exists. What is the expected number of cars that can park along that 100-unit-long circle?

Love love love, IBM for encouraging math nerdiness (-_+)

To submit an answer to June’s challenge email webmaster@watson.ibm.com

It’s a lovely problem, but I didn’t understand the so-called “solution”. Where, for example, did

f(0) = f(1) = 0, since there is not enough space for a 2-unit-long car and f(n+1) = 1 + {2\over n}\sum_{i=0}^{n-1} f(i)

…come from!?

The answer is 42 +1.

Haha, email your response to webmaster@watson.ibm.com to see if you are still eligible for the prize!