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Can You Unravel These Number Strings?

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,1 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 Peter Neuhaus, a puzzling sequence:

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
312,211

He wants to know one thing: What number comes next?

Submit your answer

Riddler Classic

From Brad Ramirez, a second shrewd sequence:

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:

This sequence creates itself, number by number, as explained in the image above. Each digit refers to the number of consecutive 3s before a certain 2 appears. 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 us questions about it. What is the ratio of 3s to 2s in the entire sequence?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Kari Matthews-Vaughn 👏 of Tulsa, Oklahoma, winner of last week’s Express puzzle!

You’re the president of the United States and you have a problem: Someone on your staff keeps leaking stories to the press. So you and your new chief of staff devise a plan. You will give different pieces of information to different staffers so that you’ll learn who the leakers are by seeing what information ends up in the newspaper or on TV. (You know you have only one leaker, and you know he or she leaks any story he or she is given.) If there are 100 people on your staff, how many different stories do you need to be able to identify your leaker for sure?

You’ll need seven stories.

The trick is to release stories sequentially. Here’s your strategy: Tell half your staff (50 people) one story and withhold it from the other half (also 50 people). If it leaks, you know the leaker is in the first half, and if doesn’t, you know the leaker is in the second half. Do the same thing again with the 50 remaining suspects: Give a second story to 25 of them and withhold it from the other 25. Once you see whether it leaks or not, you’ll have narrowed it down to 25 suspects. Rinse and repeat.

With this strategy, you’ll narrow down the staff, in the worst-case scenario, in a sequence like so: 100, 50, 25, 13, 7, 4, 2, 1. (If you get lucky, the leaker will be in one of the smaller groups created when you split the odd numbers in half, which might let you find them sooner.) So in seven steps, you’ll find the leaker for sure.

This is an example of a binary search. In the worst-case scenario, this type of search will take \(\log_2 100\) (rounded up to the nearest whole number) steps. The number \(\log_2 100\) equals about 6.6, so we’ll need 7 steps to guarantee we find the leaker.

Solution to last week’s Riddler Classic

Congratulations to 👏 Seth Colbert-Pollack 👏 of Minneapolis, winner of last week’s Classic puzzle!

Fans of “Dungeons & Dragons” will have fond feelings for four-sided dice, which are shaped like regular tetrahedrons. Some of you might have noticed, in those long hours of fantasy battle, that if you touch five of these pyramids face-to-face-to-face, they come agonizingly close to forming a closed pentagon. Alas, there remains a tiny angle of empty space left between two of the pyramids. What is the measure of that angle?

It’s about 7.4 degrees.

If you knew what to look for, you could pretty quickly work out the answer to this problem via Wikipedia. The site lists a tetrahedron’s dihedral angle — the angle between two intersecting planes — as 70.528779 degrees. There are five dice, so five such angles. There are 360 degrees in all, so \(360-5\cdot 70.528779 =\) a gap of 7.356105 degrees.

But if you eschewed help from the internet, or if you just wanted to start from scratch, here’s how you could get there, adapted from a tidy solution submitted by Matthew Baron.

Assume the sides of the triangular faces of the dice have a length of 1. Now, begin with a single face and find the height of the triangle.

We know from the Pythagorean theorem that \(h^2+0.5^2=1^2\), so \(h=\sqrt{3}/2\).

Now we want to find the angle between two faces — the dihedral angle. Let’s call it \(\gamma\).

We can find the angle \(\gamma\) using the law of cosines:

\begin{equation*}\cos(\gamma)=\frac{-1+3/4+3/4}{2\cdot 3/4}\end{equation*}

Simplifying, \(\cos(\gamma)=\frac{1}{3}\), so \(\gamma \approx 70.53\) degrees. And, again, subtracting five such angles from a total of 360 degrees (\(360-5\cdot 70.53\)) gives our answer: about 7.4 degrees.

Solver Amy Leblang happened to have some tetrahedral dice on hand, and she verified the result:

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.

Footnotes

  1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday. Have a great weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.

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