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

From John Torrey comes a friendly sort of puzzle:

John recently took an online quiz. One of the questions asked which three actors from the original “Scream” movie had returned for the fifth installment of the franchise. There were five choices:

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

*You* might know the answer. But suppose that, like John, you have no idea. If you were to randomly choose three of the five actors, what is the probability that you would get at least two right?

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

## Riddler Classic

Worried about hyperpolarization? Then this is *not* the puzzle for you.

I have a light source that’s polarized in the vertical direction, but I want it to be polarized in the horizontal direction. To make that happen I need … polarizers! When light passes through a polarizer, only the light whose polarization aligns with the polarizer passes through. When they aren’t perfectly aligned, only the *component* of the light that’s in the direction of the polarizer passes through.

Unfortunately, that means I can’t turn vertically polarized light into horizontally polarized light with a single polarizer. But I *can* do it with two polarizers, as shown below.

If the first polarizer is positioned at an acute angle with respect to the light, and then the second polarizer is positioned at a complementary angle with respect to the first polarizer, some light will be horizontally polarized in the end.

Now, I have tons of polarizers, and each one also reflects 1 percent of any light that hits it — no matter its polarization or orientation — while polarizing the remaining 99 percent of the light.

I’m interested in horizontally polarizing as much of the incoming light as possible. How many polarizers should I use?

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

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Gareth McCaughan ÑÑâÐ of Cambridge, United Kingdom, winner of last week’s Riddler Express.

Last week, you played a “game” in which you started with five shaded squares on an infinitely large grid, in the following formation:

That was generation 1. To go from one generation to the next, you had to consider every square’s eight neighbors (up, down, left, right and the four diagonal directions). If *at least three* of those squares were shaded in the previous iteration, that square was shaded in the next generation.

That said, here were the first four generations:

How many squares were shaded in generation 10?

Solver Diarmuid Early animated the first 36 generations of growth:

Between generations 1 and 2 and generations 2 and 3, four more squares were added per generation. Eight squares were added for each of the next three generations, and the three generations after that saw 12 new squares per generation. By generation 10, there were **89** shaded squares, the solution to the puzzle.

But this pattern — three generations with the same increasing multiple of four — continued ad infinitum. At first, the mass of squares looked like a growing circle. But over time, a more precise pattern became apparent. On alternating generations, new squares poked out of the top, bottom, left and right, forming cascades of newly shaded squares on either side of them. Eventually, these cascades met up halfway between each of those four corners. By generation 36, it was clear that the pattern was not circular but rather *octagonal*.

For extra credit, you had to determine approximately how many squares were shaded in generation *N*. Many solvers, like Emily Boyajian, found an exact formula. But here I’ll stick with an approximation.

After *N* generations, you had added four squares for three generations, then eight squares for another three generations, then 12, then 16 and so on. You were adding approximately *N*/3 values in triplicate, and these values were multiples of 4, from 4 to 4*N*/3. Factoring out those 4s meant the total was approximately 12 times the sum of the whole numbers from one to *N*/3, or 12·*N*/3·(*N*/3+1)/2. The leading term here, which described the number of squares when *N* was very large, was **2***N*^{2}**/3**.

So not only were these squares growing octagonally, but also quadratically. Fascinating!

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Benjamin Johnson ÑÑâÐ of White Plains, New York, winner of last week’s Riddler Classic.

Last week, you wanted to change the transmission fluid in your old van, which held 12 quarts of fluid. Initially, all 12 quarts were “old.” But changing all 12 quarts at once carried a risk of transmission failure.

Instead, you decided to replace the fluid a little bit at a time. Each month, you removed one quart of old fluid, added one quart of fresh fluid and then drove the van to thoroughly mix up the fluid. Unfortunately, after precisely one year of use, formerly fresh transmission fluid officially turned “old.”

You kept up this process for many, many years. One day, immediately after replacing a quart of fluid, you decided to check your transmission. What percent of the fluid was old?

At the beginning, all 12 quarts were old. After one month, one quart was new and the other 11 quarts were old. One month later, another new quart was added, replacing some (1/12, to be exact) of the previously new quart as well as some (again, 1/12) of the old 11 quarts. This was starting to get a little complicated.

One way to cut through the clutter was to distinguish between fluids that were brand new, one month old, two months old, etc. So after two months, one quart was brand new, 11/12 was one month old, and everything else (the remaining 121/12 quarts) was old. And after three months, one quart was new, 11/12 was one month old, and (11/12)^{2} was two months old.

This pattern continued for 12 months, at which point the system reached a stable equilibrium. The new fluid included everything from the one brand new quart to the (11/12)^{11} quarts that were 11 months old. You could compute the total amount of new fluid as the sum of a finite geometric series, which gave you (1−(11/12)^{12})/(1−11/12). Since 12 also happened to be the number of quarts (and not just the number of months that had passed), the *fraction* of fluid that was new was 1−(11/12)^{12}. The remainder had to be old, which was (11/12)^{12}, or approximately **35.2 percent**.

By the way, this percentage didn’t vary much after the first year. While 35.2 percent of the fluid was old at the beginning of each month, 38.4 percent was old at the end of each month. And once you added more fluid, it dropped right back down to 35.2 percent.

Anyway, *please* don’t apply the results of this puzzle to your own vehicle’s transmission.

## Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

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