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Can You Solve This Napoleonic Puzzle?

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world, including you, the readers. You’ll find this week’s puzzle below.

Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the form at the bottom. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible for the shoutout, I need to receive your correct answer before 11:59 p.m. EDT on Sunday — have a great weekend!

Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 Kirk Sosebee 👏 of Sautee, Georgia, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

Now here’s this week’s Riddler, short but deadly, a Napoleonic puzzle. It comes to us from Andy Laursen of San Francisco.


Complete this series:

10, 11, 12, 13, 14, 15, 16, 17, 21, 23, 30, 33, …

(Yep, that’s it.)


Need a hint? You can try asking me nicely. Want to submit a puzzle or problem? Email me.


And here’s the solution to last week’s Riddler, concerning a number-guessing game played by some expert logicians. I received the most submissions ever for this problem. Well done, Riddler Nation! Not only that, but 84.8 percent of them were correct.

The pair of numbers Barack was thinking of are 2 and 8. Here’s one way to arrive at them: Make a grid with the numbers 1 through 9 as rows and 1 through 9 as columns. So now, each entry in this grid is a possible pair of numbers in Barack’s head. (You can eliminate a big triangular chunk of the grid since 3 and 5, say, is the same pair as 5 and 3.) Then, begin eliminating entries. For example, when Pete first says he doesn’t know what the numbers are, we can eliminate all the entries with unique products, since Pete knows how to do simple math. The pair can’t be 1 and 2, for example, since that’s the only pair with a product of 2. If the product of the numbers were unique, then Pete would’ve known the pair. Then do the same for the sums when it’s Susan’s turn, and then the products, and then the sums, and then the products, and so on. Once we get to Pete’s fifth turn, he says he knows the pair, and the only pair left with a unique product at that point is 2 and 8.

Here’s a good illustration of this process from Joel Baker:

baker

As usual, I received a bountiful bevy of beautiful submissions. Would that we could include them all, but here’s just one more, a lovely analog job from Irina Smoke:

Lastly, in an epic tweet that should by rights have about 50,000 retweets by now, Riddler Hall of Famer Zach Wissner-Gross explored what happens when you expand the possible set of Barack’s numbers beyond just 1 through 9. So many patterns! Buried in here, I’m absolutely convinced, is the meaning of life:

From all of us here at Riddler Headquarters: Have a wonderful weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.

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