Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either, and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From Alex Vornsand, a puzzle for your daily routine:

You take half of a vitamin every morning. The vitamins are sold in a bottle of 100 (whole) tablets, so at first you have to cut the tablets in half. Every day you randomly pull one thing from the bottle — if it’s a whole tablet, you cut it in half and put the leftover half back in the bottle. If it’s a half-tablet, you take the vitamin. You just bought a fresh bottle. How many days, on average, will it be before you pull a half-tablet out of the bottle?

Extra credit: What if the halves are less likely to come up than the full tablets? They are smaller, after all.

Submit your answer

Riddler Classic

From Mikael Rittri, a mathematical souvenir problem:

In the Riddler gift shop, we sell interesting geometric shapes of all sizes — Platonic solids, Archimedean solids, Klein bottles, Gabriel’s horns, you name it — at very fair prices. We want to create a new gift for fall, and we have a lot of spheres, of radius 1, left over from last year’s fidget sphere craze, and we’d like to sell them in sets of four. We also have a lot of extra tetrahedral packaging from last month’s Pyramid Fest. What’s the smallest tetrahedron into which we can pack four spheres?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Remy Cossé 👏 of Philadelphia, winner of last week’s Express puzzle!

A mysterious figure emerges from the shadows and hands you a note with the following list of six numbers: 1, 11, 21, 1,211, 111,221 and 312,211. He wants to know one thing: What number comes next?

It’s 13,112,221.

This sequence is sometimes called the “say what you see sequence.” You start with a single number — 1 — and then to get the next number you say what you see. First, you see one 1, so you write down 11. Then, you see two 1s, so you write down 21. Then you see one 2 and one 1, so you write down 1,211. To get the number requested by the shadowy figure, you say what you see one last time: one 3, one 1, two 2s and two 1s, or 13,112,221.

Solution to last week’s Riddler Classic

Congratulations to 👏 Cody Couture 👏 of Irvine, California, winner of last week’s Classic puzzle!

Take a look at this string of numbers: 333 2 333 2 333 2 33 2 333 2 333 2 333 2 33 2 333 2 333 2 … At first it looks like someone fell asleep on a keyboard. But there’s an inner logic to the sequence: It creates itself, number by number. Each digit refers to the number of consecutive 3s that appear before a certain 2. Specifically, the first digit refers to the number of consecutive 3s that appear before the first 2, the second digit refers to the number of 3s that appear consecutively before the second 2, and so on toward infinity. The sequence never ends, but that won’t stop us from asking questions about it. What is the ratio of 3s to 2s in the entire sequence?

It’s \(1+\sqrt{3}\), or about 2.73-to-1.

Solver Adam Palay approached the problem by thinking of the numbers in terms of “generations.” A first-generation 3 spawns a second-generation 3332, for example, while a first-generation 2 spawns a second-generation 332.

With that in mind, Palay explained, we can set up some equations. Let each generation be denoted by a number n, and let “threes” and “twos” be the number of 3s and 2s in that generation. Those numbers evolve like this:

\begin{equation*}threes_{n+1} = 3\cdot threes_n + 2 \cdot twos_n\end{equation*}

\begin{equation*}twos_{n+1} = threes_n + twos_n\end{equation*}

So now we need to get from that evolution to the overall ratio. Call that ratio r.

\begin{align*} r_{n+1} &= \frac{threes_{n+1}}{twos_{n+1}} = \frac{3\cdot threes_n + 2 \cdot twos_n}{threes_n +twos_n} \\&= 2 + \frac{threes_n}{threes_n+twos_n}\\&= 2 + \frac{1}{1+\frac{twos_n}{threes_n}}\\&= 2 + \frac{1}{1+\frac{1}{\frac{threes_n}{twos_n}}} \end{align*}

And \(\frac{threes_n}{twos_n}\) is exactly \(r_n\). So we arrive at the following:

\begin{equation*}r_{n+1} = 2 + \frac{1}{1+\frac{1}{r_n}}\end{equation*}

We can also guess that as the sequence grows and many, many generations are spawned, \(r_n\) and \(r_{n+1}\) converge to the same number — the single ratio the puzzle was asking for. (Hector Pefo provided some more explanation of this part of the solution.) So all that’s left is to solve the following for r:

\begin{equation*}r = 2 + \frac{1}{1+\frac{1}{r}}\end{equation*}

A little algebra gives the solution: \(r = 1 + \sqrt{3}\).

Solver Luke Benz showed empirically how quickly the ratio of 3s to 2s converges to about 2.73:

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.