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-size 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 have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Tyler Barron, you spin me right round, numbers, right round:

Given a two-character, seven-segment display, like you might find on a microwave clock, how many numbers can you make that are *not* ambiguous if the display happens to be upside down?

For example, the number 81 on that display would not fit this criterion — if the display were upside down it’d appear like 18. The number 71, however, would be OK. It’d appear something like 1L — not a number.

## Riddler Classic

Crossword puzzle grids typically obey a few rules and conventions.

- They are 15-by-15.
- They are rotationally symmetric — that is, if you turn the grid upside down it appears exactly the same.
- All the words — that is, all the horizontal and vertical sequences of white squares — must be at least three letters long. All the letters must appear in an “across” word and a “down” word.
- The grid must be entirely connected — that is, there can be no “islands” of white squares separated from the rest by black squares.

First question: How many such crossword grids are there?

Second question: Crossword constructors do well to avoid using “cheater squares,” black squares whose addition makes some words shorter but does not change the puzzle’s total word count. How many grids are there without cheater squares?

*Extra credit: *The Sunday “New York Times” puzzle is 21-by-21. How many of those are there, with and without cheater squares?

## Solution to the previous Riddler Express

Congratulations to 👏 Grace Lyden 👏 of Minneapolis, winner of last week’s Riddler Express!

Last week, you multiplied together some of the integers from 1 to 99 and got this monstrosity as a result:

530,131,801,762,787,739,802,889,792,754,109,70_,139,358,547,710,066,257,652,050,346,294,484,433,323,974,747,960,297,803,292,989,236,183,040,000,000,000.

What was the missing digit?

It was **6**.

The puzzle’s submitter, Max Weinreich, explains the logic: Hopefully, this number is divisible by 9. If it is, then we know that its digits add up to a multiple of 9, which, upon adding all the digits up, would force the missing digit to be a 6. How can I be sure that my number really is divisible by 9? The largest number that is a product of integers from 1 to 99 but that is *not* divisible by 3 is 1 * 2 * 4 * 5 * 7 * 8 * 10 * 11 * … * 97 * 98. And so any two extra numbers I throw into this calculation will force the answer to be a multiple of 3 twice over — that is, a multiple of 9. So the largest non-multiple of 9 that I could get by my calculation is 96 * (1 * 2 * 4 * 5 * … * 97 * 98) which, if you put it into the calculator, turns out to have fewer digits than the enormous number in our problem. Therefore, the number is divisible by 9 and the missing digit is a 6.

## Solution to the previous Riddler Classic

Congratulations to 👏 Viviana Acquaviva 👏 of Brooklyn, New York, winner of last week’s Riddler Classic!

Last week you were selected for the first manned mission to Mars: five Earth-years, or 1,825 Earth-days, on the red planet. Conditions would be brutal. So brutal that it was known exactly one vital piece of equipment would break each day. Therefore, you and the rest of the international team were sent with three 3D printers to print replacement parts for critical equipment. Each printer was manufactured in a different country, however, and parts from one printer were not compatible with any of the other printers (that meant no scavenging allowed).

If something broke on a 3D printer, you had to use one of the other 3D printers to print a replacement part. Any part could be printed effectively instantly, though any given printer only had the power to print one piece per day. The Riddler Aeronautics and Space Administration tested all three printers and found that, in addition to the daily breakage of the vital life-support equipment, one had a 10 percent chance of something breaking on any given day, the second a 7.5 percent chance and the last a 5 percent chance. If you couldn’t quickly print a replacement part for any piece of vital equipment, you’d die. What were the chances you made it home alive?

If you were judicious in your use of the printers, you’d have about a **50-50 shot** at survival. Let’s walk through some interplanetary disaster scenarios.

If *zero printers* break on some day, you’re clearly fine. You have more than enough hardware to fix your vital life-support equipment. You could even create some Martian art with your extra 3D printers.

If *one printer* breaks on some day, you’re also fine. You can print a part to fix it with one of the printers, and you can print a part to fix the vital equipment with the other printer. No sweat. (After all, it is about -80 degrees Fahrenheit up there.)

If *two printers* break on some day, you may begin to worry. However, you can science your way out of this scenario, too. Call the printers A, B and C, and suppose it was A and B that broke. Fix A using C. Then fix B using A. Then fix the vital equipment using B. Phew! Close one.

If *three printers* break on some day, you’re toast. Or an icicle. In any case, you’re dead — you have no way to fix the vital equipment that breaks. The chances of all three printers breaking on a given day are \(0.1\times 0.075 \times 0.05 = 0.000375\), or 0.0375 percent. Not bad, but you’re going to be on Mars for a while — 1,825 days to be precise. The chance that this deadly state of affairs never comes to pass is \((1-0.000375)^{1,825}\), or about 0.5043, or about **50.43 percent**.

Would you flip a life-and-death coin for five years on Mars? (I would.) Or would you just find a way to decommission Riddler Nation’s space agency for putting you in such unnecessary peril in the first place?

## 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 me at oliver.roeder@fivethirtyeight.com.

**CORRECTION (Jan. 18, 2018, 10:40 a.m.)**: An earlier version of this article contained a logically incomplete Riddler Express solution. That solution has been replaced.