Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the 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 the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

Trapezoidal numbers are whole numbers that can be written as the sum of two or more consecutive positive integers. For example, 9 is a trapezoidal number because it can be written as 2+3+4. In fact, 9 is *doubly *trapezoidal because it can also be written as 4+5. (By the way, triangular numbers are also trapezoidal. Iām a fan of inclusive definitions like these.)

What is the smallest number that is *triply* trapezoidal, meaning it is trapezoidal in three different ways?

*Extra credit: *What is the smallest number that is *quadruply* trapezoidal? *Quintuply* trapezoidal? *Sextuply* trapezoidal?Ā

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

From Dean Ballard comes a cubical conundrum:

Consider a cube, which has eight vertices, or corners. Suppose I assign a prime number to each vertex. A āface sumā is the value I get when I add up all four prime numbers on one of the six faces.

Can you find eight distinct primes and arrange them on a cube so that the six face sums are all equal?

*Extra credit:* Can you find another set of eight distinct primes that can similarly be arranged on the vertices of a cube? How many more can you find?

The solution to this Riddler Classic can be found in the following column.

## Solution to last weekās Riddler Express

Congratulations to š Malcolm Gray š of Cambridge, winner of last weekās Riddler Express.

Last week, you analyzed an online quiz that asked which three actors from the original āScreamā movie had returned for the fifth installment of the franchise. There were five choices:

- Courteney Cox
- David Arquette
- Halle Berry
- Neve Campbell
- Tom Holland

For the sake of the problem, you supposed that you had no clue about any of these actors.

If you randomly chose three of the five actors, what was the probability that you would have gotten at least two right?

One way to figure this out was to first compute the total number of ways to choose three of the actors. Since the order of your selections didnāt matter, this was 5 choose 3, or 10.

Of those 10 ways to choose three actors, only one of them got all three correct. But how many ways were there to get precisely *two* correct? There were three different pairs of correct actors you could have identified (Cox and Arquette, Cox and Campbell, and Arquette and Campbell). For each of those three pairs, there were two incorrect actors to choose from (Berry and Holland). In other words, there were 3 times 2, or six, ways to get exactly two actors right.

With seven ways to correctly identify *at least* two actors out of a 10 ways to choose any three, the answer was **70 percent**.

Another approach, used by the students in The Problem Solving & Posing Class at New York Cityās Hewitt School, was to count up how many of the 10 cases resulted in you getting only one actor correct. (Note that you couldnāt get all three wrong.) You could choose any one of the three correct actors, and then you had to select both incorrect actors along with them. So of the 10 ways to choose three actors, three ways resulted in you *not* getting at least two right. Once again, the answer was 70 percent.

Finally, I will say that several readers, including Amy from Wayland, Massachusetts and Rick Wingo from Austin, Texas, pointed out that Tom Holland was born only months before the release of the original Scream film. He clearly was not a correct choice. But now I canāt get the image of a newborn Tom Holland in a Ghostface costume out of my head.

## Solution to last weekās Riddler Classic

Congratulations to š Colin Parker š of Portola Valley, California, winner of last weekās Riddler Classic.

Last week, I had a light source that was polarized in the vertical direction, but I wanted it to be polarized in the horizontal direction. To make that happen I needed ā¦ polarizers! When light passes through a polarizer, only the light whose polarization aligns with the polarizer passes through. When they arenāt perfectly aligned, only the *component* of the light thatās in the direction of the polarizer passes through.

Unfortunately, that meant I couldnāt turn vertically polarized light into horizontally polarized light with a single polarizer. But I *could* do it with two polarizers, as shown below.

If the first polarizer was positioned at an acute angle with respect to the light, and then the second polarizer was positioned at a complementary angle with respect to the first polarizer, some light would have been horizontally polarized in the end.

Now, I had tons of polarizers, and each one also reflected 1 percent of any light that hit it ā no matter its polarization or orientation ā while polarizing the remaining 99 percent of the light.

I wanted to horizontally polarize as much of the incoming light as possible. How many polarizers should I have used?

First of all, how much light could you horizontally polarize with *two* polarizers? If the first polarizer was an angle š from vertical, then the second polarizer had to be 90Ā°āš from the first polarizer. Assuming no reflection (for now), that meant the component of the light getting through was cos(š)Ā·cos(90Ā°āš). This function was maximized when š was 45Ā°, meaning the two polarizers evenly split the 90Ā° angle into two 45Ā° angles.

That was when you had two polarizers. What about *N* polarizers? Once again, your best strategy is to evenly divide the 90Ā° angle into *N *equal angles, each measuring 90Ā°/*N*. Several solvers went on to prove this in different ways: Laurent Lessard with Jensenās inequality, Rajeev Pakalapati with induction and Arvind Hariharan with a symmetry argument. With *N* evenly spaced polarizers, the final horizontal component was cos* ^{N}*(90Ā°/

*N*).

Now before we go any further, letās say a little more about light. Up to this point, the vectors Iāve referred to are the electric field vectors of the light. However, the *intensity* of the light is proportional to the *square* of the magnitude of this electric field vector. When it came to positioning the polarizers, this didnāt matter ā maximizing the electric field was effectively the same as maximizing its square. However, this ambiguity affected your interpretation of the fact that 1 percent of the light was reflected.

Most readers interpreted this as meaning that the electric field vector attenuated by 1 percent with each polarizer. In this case, the final horizontal component was (0.99)* ^{N}*Ā·cos

*(90Ā°/*

^{N}*N*). Thanks to the exponent on the cosine function, this function does some strange things when

*N*is less than 1. But when

*N*was greater than 1, the maximum occurred when

*N*was approximately 11.135. Since you had to have an integer number of polarizers, the optimal number turned out to be

**11**, in which case the horizontal component was about 80 percent of the original vertical component.

Other readers interpreted the puzzle as saying the lightās *intensity* was attenuated by 1 percent with each polarizer. Since the intensity was proportional to the square of the electric field, this meant the final horizontal component was (0.99)^{N}^{/2}Ā·cos* ^{N}*(90Ā°/

*N*). This time around, the maximum occurred when

*N*was approximately 15.708. The optimal (integer) number of polarizers was

**16**, in which case the horizontal component was about 85.42 percent of the original vertical component.

If you thought two interpretations of the puzzle werenāt enough, then worry not. Solver Joao Coelho had a *third* interpretation, where the light that wasnāt transmitted by each polarizer was reflected back to the previous polarizer, and could bounce back and forth internally multiple times. According to Joao, accounting for these internal reflections allowed you to squeeze in one more polarizer.

With all these possible interpretations, this puzzle turned out to be a little more polarizing than I expected.

## Want more riddles?

Well, arenāt you lucky? Thereās a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. Itās called āThe Riddler,ā and itās in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.