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
Earlier this year, a new generation of Brood X cicadas had emerged in many parts of the U.S. This particular brood emerges every 17 years, while some other cicada broods emerge every 13 years. Both 13 and 17 are prime numbers — and relatively prime with one another — which means these broods are rarely in phase with other predators or each other. In fact, cicadas following a 13-year cycle and cicadas following a 17-year cycle will only emerge in the same season once every 221 (i.e., 13 times 17) years!
Now, suppose there are two broods of cicadas, with periods of A and B years, that have just emerged in the same season. However, these two broods can also interfere with each other one year after they emerge due to a resulting lack of available food. For example, if A is 5 and B is 7, then B’s emergence in year 14 (i.e., 2 times 7) means that when A emerges in year 15 (i.e., 3 times 5) there won’t be enough food to go around.
If both A and B are relatively prime and are both less than or equal to 20, what is the longest stretch these two broods can go without interfering with one another’s cycle? (Remember, both broods just emerged this year.) For example, if A is 5 and B is 7, then the interference would occur in year 15.
The solution to this Riddler Express can be found in the following column.
Riddler Classic
The astronomers of Planet Xiddler are back!
This time, they have identified three planets that circularly orbit a neighboring star. Planet A is three astronomical units away from its star and completes its orbit in three years. Planet B is four astronomical units away from the star and completes its orbit in four years. Finally, Planet C is five astronomical units away from the star and completes its orbit in five years.2
They report their findings to Xiddler’s Grand Minister, along with the auspicious news that all three planets are currently lined up (i.e., they are collinear) with their star. However, the Grand Minister is far more interested in the three planets than the star and wants to know how long it will be until the planets are next aligned.
How many years will it be until the three planets are again collinear (not necessarily including the star)?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Paul Juster 👏 of Walnut Creek, California, winner of last week’s Riddler Express.
Last week, you studied a full, 200-episode season of Riddler Jeopardy!. The first episode of the season featured three brand-new contestants. Each subsequent episode included a returning champion (the winner of the previous episode) as well as two new challengers.
Throughout the season, it so happened that the returning champions were particularly strong, with each one winning five consecutive episodes before being dethroned on the sixth.
If you picked a contestant at random from the season, what was the probability that they were a Riddler Jeopardy! champion (meaning they won at least one episode)?
To figure this out, you had to determine both the total number of champions in a season as well as the total number of contestants. The solution was then the former divided by the latter.
The number of champions wasn’t too tough. There were 200 episodes, with each champion winning five episodes. That meant there were 200 divided by 5 — or 40 — champions.
But what about the total number of contestants? Counting them up in games with new champions as opposed to returning champions was, for many readers, rather tricky.
Solver Jesus Petry of Porto Alegre, Brazil, worked this out in a straightforward way. Jesus knew that regardless of who was the winner, every episode introduced two new contestants. Every episode, that is, except the first, which introduced three new contestants. That meant the total was 199·2+3, or 401.
With 401 total contestants and just 40 champions, your chances of picking a winner were 40/401, or about 9.975 percent.
What does this mean? Well, you might intuitively think that a third of Jeopardy! contestants are winners, since each episode has three contestants and one of them wins. However, because winners come back to play in the next episode, that effectively reduces how many winners there are.
But as unlikely as it might be to become a Jeopardy! champion, it’s still better than your chances of being the next host.
Solution to last week’s Riddler Classic
Congratulations to đź‘Ź Peter Exterkate đź‘Ź of Sydney, Australia, winner of last week’s Riddler Classic.Â
On June 24, the following challenge was posted on Twitter:
In this challenge, the numbers from 1 to 11 were arranged in a circle in a particular order: 1, 4, 8, 7, 11, 2, 5, 9, 3, 6, 10. You then had to connect pairs of numbers with straight line segments that didn’t intersect, and your score was the sum of the products of the joined numbers. For example, with the connections {1, 4}, {8, 10}, {3, 7}, {5, 9}, and {2, 11} (and the 6 left by itself), you got a score of 1·4 + 8·10 + 3·7 + 5·9 + 2·11, or 172.
The best score you could achieve with this ordering of 1 through 11 around the circle was 237, which you could get with the following connections: {6, 10}, {3, 4}, {7, 8}, {9, 11} and {2, 5} (and the 1 left by itself).
But what if you wanted the lowest possible maximum score? That is, how could you have ordered the numbers from 1 to 11 around the circle so that the maximum possible score was as low as possible? And what was the resulting score?
This puzzle was doubly challenging. For any given permutation of numbers around the circle, finding the maximum score was hard enough. But on top of that, you also had to find the arrangement that resulted in the smallest maximum score. I can report that everyone who found the correct answer did so with the assistance of a computer and by checking all possibilities.
Without further ado, here was one such optimal arrangement:
Before reading on, if you haven’t completely solved the puzzle yet, you can still take a crack at it! Yes, this is an optimal arrangement, but what’s the highest score you can get with it? Read on for the complete solution …
The most points you could squeeze out of the above arrangement was 224. As it turned out, there were two different ways to do it. Here was one way:
And here was another:
As I said, this was one optimal arrangement. Peter, as well as Emily Boyajian of Minneapolis, Minnesota, found another 21 (for a total of 22) that were all distinct, meaning no two were rotations or reflections of each other.
Emily went one step further, studying how likely different maximum point totals were. The lowest such score was 224, while the highest was 250. With random arrangements of the 11 numbers around the circle, it turned out that higher-scoring totals were more likely:
Personally, I found it rather surprising that when all was said and done, the range of maximum scores only varied by about 10 percent. Now if only my bowling scores were this consistent.
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.