# Some questions from the Museum of Math’s fundraiser

Numberplay: Math as a spectator sport. From Numberplay, a sub-blog of Wordplay, the crossword blog of the New York Times. The Museum of Mathematics, which is supposed to open later this year, held a fundraising event which included a mathematical tournament. (MCed by Will Shortz! I’m kind of curious how much math he knows.)

Go there and look at the questions; I’m going to solve some of them here, with my own comments. (In the contest these were multiple-choice. Since multiple-choice questions feel wrong outside of tests, and this blog is not a test, I’m omitting the choices.)

Moon creatures, let us suppose, have a unit of distance that we shall call a “lunar”. It was adopted because the moon’s surface area, if expressed in square lunars, exactly equals the moon’s volume in cubic lunars. The moon’s diameter is 2,160 miles. How many miles long is a lunar?

This is actually an interesting way to measure area. We’ll figure out how many lunars there are in a moon-radius. Since the surface area has to equal the volume, we get 4πr2 = (4/3)πr3. Solving for r gives r = 3, so there are three lunars in a moon-radius, or six in a moon-diameter. The answer is 360. (The moon then has area and volume 36π.)

This feels roughly analogous to the definition of “radian” for circles, but is not quite the same. In fact, solving a similar equation for circumference = area, 2πr = πr2, we get r = 2, suggesting we should take half the radius of a circle as the natural measure and giving 4π for both the circumference and the area. (The Tauists might be glad to hear that.)

Call a set of integers “spacy’ if it contains no more than one out of any three consecutive integers. How many subsets of {1, 2, 3, ……,12} are spacy?

Let g(n) be the number of spacy subsets of $\{1, 2, ..., n\}$. Then clearly $g(0) = 1, g(1) = 2, g(2) = 3, g(3) = 4$ — in each case the spacy sets are just the empty set and the one-element subsets.

Then for larger n, $g(n) = g(n-1) + g(n-3)$. How can you make a spacy subset of $\{1, 2, \ldots, n\}$? Well, any spacy subset of $\{ 1, 2, \ldots, n-1 \}$ is a spacy subset of $\{ 1, 2, \ldots, n \}$ which does not contain \$lates n\$, and there are $g(n-1)$ of those. And any spacy subset of $\{ 1, 2, \ldots, n \}$ which contains \$n\$ must, when you remove $n$ from it, give a spacy subset of $\{ 1, 2, \ldots, n-3 \}$. There are \$g(n-3)\$ of those. Any spacy subset of \$\latex \{1, 2, \ldots, n\}\$ either contains or does not contain $n$, and so $g(n) = g(n-1) + g(n-3)$.

For example, the spacy subsets of \$\{ 1, 2 \}\$ are $\emptyset, 1$, and \$2\$ (for notational ease I’m dropping the braces\$. The spacy subsets of $\{ 1, 2, 3, 4 \}$ are $\emptyset, 1, 2, 3, 4, 14$. So the spacy subsets of $\{ 1, 2, 3, 4, 5 \}$ are \$\emptyset, 1, 2, 3, 4, 14\$ (from the spacy subsets of $\{ 1, 2, 3, 4\}$) and $5, 15, 25$ (from adding a 5 to each spacy subset of $\{ 1, 2\}$.)

Interestingly, the winners in this contest all came out of the financial community: the affiliations mentioned in the post at the Times are Two Sigma Investments, Renaissance Technologies, Morgan Stanley, and Citadel. A few explanations for this present themselves:

• people who can solve these sorts of problems quickly tend to end up in that industry;
• this was a fundraising event and people in finance have lots of money;
• this event took place in New York; you’d have different results if you tried this in, say, San Francisco

The last of these explanations would be easy to check.

## One thought on “Some questions from the Museum of Math’s fundraiser”

1. Mark Dominus says:

I used to know John Overdeck (the contest winner), although I haven’t seen him since 1984. At that time, he was training himself to defeat the world record mental arithmetic. The problem was described as extracting the 13th root of a 100-digit number, but it was actually somewhat easier than this; the 100-digit number was guaranteed to be a perfect 13th power. So the result must be an 8-digit integer, and the first and last digits are trivial. Overdeck had come up with various arithmetic tricks to find the other 6 digits, the most complicated one involved a 2×4 mental multiplication, and then a table lookup on table with (I think) a few hundred entries, enough to feasibly memorize.

Overdeck was also a silver medalist at the 1986 IMO.