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.
We have another Riddler video! This episode is about RoboPizza, where robots cut your pizza. Watch me walk through the steps to determining how many pieces you can expect from the robot chefs.
The Flash challenges Usain Bolt to a 100-meter race. Bolt runs at an average speed of 10 meters per second. To make it interesting, the Flash decides he will pick a random speed between 5 meters per second and 16 meters per second, with each speed in between being equally likely. (Note that fractional and decimal speeds are included here, rather than just whole numbers.)
On average, how often would you expect the Flash to win? What would be his average margin of victory?
From Jenny Mitchell comes a matter of shoring up your defenses:
In the hit online game World of Riddlecraft, players can level up their armor. Armor levels range from 0 to 5. Now, attempting to level up your armor requires a cerulean gem, which is destroyed in the process. If the attempt is successful, your armor’s level goes up by one; if not, it goes down by one.
Fortunately, it’s impossible to fail when attempting to upgrade your armor from level 0 to level 1. However, the likelihood of success goes down the higher level the armor is before the upgrade. More specifically:
- Upgrading from level 0 to level 1 has a 100 percent chance of success.
- Upgrading from level 1 to level 2 has an 80 percent chance of success.
- Upgrading from level 2 to level 3 has a 60 percent chance of success.
- Upgrading from level 3 to level 4 has a 40 percent chance of success.
- Upgrading from level 4 to level 5 has a 20 percent chance of success.
On average, how many cerulean gems can you expect to use up in order to upgrade your armor from level 0 to level 5?
Solution to last week’s Riddler Express
Congratulations to 👏 Nicholas Etz 👏 of Sarasota, Florida, winner of last week’s Riddler Express.
Last week, you had to find a word in the English language to which you could add a vowel, resulting in another word that had fewer syllables.
By “add a vowel,” I meant you had to insert one additional letter — a vowel — somewhere in the word (or at the beginning or end) while keeping the ordering of all the other letters the same. For example, you could have added a vowel to the word TASTY to get the word TOASTY. However, both words were two syllables, meaning this was not the solution.
Several readers suggested the answer was adding a T to BOA to get BOAT. While BOAT had fewer syllables than BOA, T was not a vowel. So that was not a solution.
I did receive many words that were kind of, sort of, maybe correct, but not particularly satisfying answers for one reason or another. Add an A to two-syllable (or is it just one?) FIRE to get the archaic spelling FAIRE, or to two-syllable (or is it just one again?) RILED to get RAILED. Add an O to COED (which is either a noun that’s dated, or an adjective that’s more or less an abbreviation for “co-educational”) to get COOED. Add an A to OK (is that even a word?) to get OAK.
Before you write to me to complain, I’m not saying these submissions were wrong. But there were others I liked more. Here were a few of my favorites:
- Many solvers, like Nicholas, added a U to four-syllable BEATIFY to get three-syllable BEAUTIFY.
- Several solvers, including Ava Hadaway, added a U to three-syllable CATION to get CAUTION.
- Seth Cohen and Ethan Rubin added an E to two-syllable REFED to get one-syllable REEFED.
- Others, like one fellow by the name of Oliver Roeder, added an E to two-syllable FUGU (which I just learned is a kind of fish) to get FUGUE.
I hope you enjoyed this brief foray into word puzzles. If you have a favorite, feel free to send it my way. But as always, proceed with cation.
Solution to last week’s Riddler Classic
Congratulations to 👏 Ben Knox 👏 of Madison, Wisconsin, winner of last week’s Riddler Classic.
Last week’s puzzle was all about Carmichael numbers. A positive integer N is a Carmichael number if (1) it is composite and has no square factors, and (2) if one less than every prime factor of N is a factor of N−1.
That was a lot to take in at once, so I offered an example. The smallest Carmichael number is 561. Sure enough, it has no square factors (other than 1, which we’re not counting here). Meanwhile, its prime factors are 3, 11 and 17. If we subtract one from each of those, we get 2, 10 and 16, all of which divide 560 (one less than the original number).
You were asked to find a six-digit Carmichael number that could be written as ABCABC in base 10. (Note: The digits A, B and C were not necessarily distinct.) Of course, you could have looked this up in a matter of seconds or even written some code to figure it out. But you were specifically challenged to complete this riddle by hand.
First, you might have noticed that no matter what the digits A, B and C were, ABCABC could always be factored into ABC×1,001. Furthermore, 1,001 was equal to 7×11×13, which meant all three of those were also factors of ABCABC.
So, for ABCABC to be a Carmichael number, that meant ABC couldn’t have any factors of 7, 11 or 13 (otherwise ABCABC would have a square factor). It also meant that one less than each of these numbers (6, 10 and 12) all had to divide ABCABC−1. In other words, ABCABC was congruent to 1 (mod 6), 1 (mod 10) and 1 (mod 12) — or, combining these conditions, 1 (mod 60).
We already said that ABCABC could be written as ABC×1,001, and 1,001 is congruent to 41 (mod 60). The only way for ABC×1,001 to be congruent to 1 (mod 60) was if ABC was also congruent to 41 (mod 60). (Interestingly, it so happens that 41 (mod 60) is its own modular inverse!)
There are only 15 three-digit numbers that are congruent to 41 (mod 60), and which are therefore candidates for ABC. They are: 101, 161, 221, 281, 341, 401, 461, 521, 581, 641, 701, 761, 821, 881 and 941. Recall that ABC couldn’t have any factors of 7, 11 or 13, which rules out three candidates: 161 (7×23), 221 (13×17) and 341 (11×31). The remaining 12 candidates were all prime. But we still needed ABC−1 to evenly divide ABCABC−1.
Rather than test all 12 numbers, there was a shortcut, as noted by solvers Awang B. Brantas and Emma Knight. Since all 12 candidates ended with a 1 (i.e., the digit C had to be 1), ABC−1 was simply AB0, and ABCABC−1 was AB1AB0. Since AB0 definitely divided AB0AB0, then for AB0 to also divide AB1AB0, it must have divided their difference, which was 1,000. The only candidate that was one more than a factor of 1,000 was the first number in the list: 101. Therefore, the Carmichael number in question was 101,101. Sure enough, its factors are 7, 11, 13 and 101, while 6, 10, 12 and 100 are all factors of 101,100.
Anyway, congratulations to everyone who didn’t just look up the sequence of Carmichael numbers. You did Riddler Nation proud!
Want more puzzles?
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 email@example.com.