Can You Outwit The Tax Collector?


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, 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.

Due to the holidays, the next column will appear on Jan. 7, 2022. See you in the new year!

Riddler Express

Reader Betts Slingluff enjoys holiday cryptarithms with the family and suggested that now was a good time for such a puzzle on The Riddler. This week’s Express is a spin on a cryptarithm originally by Frank Mrazik:


As with any cryptarithm, each letter represents one of the digits from 0 to 9, and different letters represent different digits.

The catch? This puzzle has two possible solutions — that is, two distinct sets of letter-to-number assignments. Can you find both solutions?

Submit your answer

Riddler Classic

This week’s Classic is an extension of a puzzle originally by Dan Finkel (and relayed to me by Fawn Nguyen):

In the game of “tax collector,” there are paychecks with different whole number dollar amounts, 1 to N. You choose a paycheck, one at a time. For any check you choose, the tax collector immediately takes any remaining checks (i.e., not already taken by you or the tax collector) whose dollar values are factors of the one you chose. For example, if you choose the $10 check, then the tax collector will immediately take $1, $2 and $5 checks — if any of them is available. Importantly, the tax collector must get something for each paycheck you choose. So if the $10 check is available, but the $1, $2 and $5 checks are not, then you cannot take the $10 check. When there are no more checks you can take, the game is over and all remaining checks go to the tax collector.

In the original version of the puzzle, your goal was tsino make more money than the tax collector when N was 12 (or 24 or 48). When N was 12, you could make $50 to the tax collector’s $28, meaning you won about 64 percent of the total. When N was 24, you could win about 61 percent of the total, and when N was 48, you could win about 62 percent.

For this puzzle, not only do you want to get more money than the tax collector, you also want to win the biggest possible fraction of the available money. Which value of N (greater than 1) would you choose, so that you can win the greatest fraction of available money?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Rick Kneedler 👏 of Portland, Oregon, winner of last week’s Riddler Express.

Last week, you were tasked with painting a building that was shaped like a regular tetrahedron. When the building was viewed from above, the architect wanted it to appear as four congruent equilateral triangles — one central blue triangle surrounded by three white triangles.

That meant that three faces of the tetrahedron contained a blue kite, as shown in the animation below:

Rotating tetrahedron. From above, it appears as an equilateral triangle whose middle quarter is blue. From the side, we can see that each of the lateral three faces has a blue kite.

What was the measure of the smallest angle in this kite?

It helped to first realize that the fraction of the total area that was blue was the same for each face whether you were looking at it from above or straight on. (For a great explanation of why that is, check out 3Blue1Brown’s recent video.) In other words, one-quarter of each of the three faces was painted blue, and the kites met the lateral edges of the tetrahedron three-quarters of the way up.

The answer didn’t depend on the overall scale of the tetrahedron, so let’s say it had side length 4, as shown in the diagram below. That meant the two top edges of the kite had length 1, while the half-width of the kite was 0.5. The height of the kite was the same as the height of the equilateral triangle, or 2√3. Again, three-quarters of that height was below the kite’s half-width, for a total length of 1.5√3.

One of the lateral faces of the tetrahedron, with labeled side lengths.

Finally, this was enough to determine half of the kite’s bottom angle. It formed a right triangle whose opposite side was 0.5 and whose adjacent side was 1.5√3. That meant the tangent of this angle was 1/(3√3), and so the angle was tan-1(1/(3√3)), or about 10.89 degrees. Again, since this was half of the bottom angle, the bottom angle was approximately 21.79 degrees.

Unlike their 2D cousins, regular tetrahedra have all sorts of angles that aren’t 30 degrees or 60 degrees, don’t they?

Solution to last week’s Riddler Classic

Congratulations to 👏 David Cohen 👏 of Silver Spring, Maryland, winner of last week’s Riddler Classic.

Last week, you were trying to whack a mole. Every second, the mole popped its head out from one of 100 holes arranged in a line. You didn’t know where the mole started, but you did know that it always moved from one hole to an adjacent hole each second. For example, if it popped out of the 47th hole, then one second later it popped out of either the 46th hole or the 48th hole. If it popped out of the first hole, then it was guaranteed to pop out of the second hole next; similarly, if it popped out of the 100th hole, then it was guaranteed to pop out of the 99th hole next.

Of course, there was a catch — the mole was camouflaged, and you had no idea where it was at any time until you actually whacked it. Each second, you could whack one hole of your choosing. Many sequences of whacks did not guarantee that you’d eventually get the mole (e.g., always whacking the first hole), but some did.

What was the shortest such sequence that guaranteed you could whack the mole, no matter where it started or how it moved around?

Before we deal with 100 holes, let’s consider a simpler case: four holes labeled 1, 2, 3 and 4. Suppose you first whacked Hole 1 and missed. Then the mole had to be in Holes 2, 3 or 4. On the next turn, after moving one hole left or right, the mole could have been in Holes 1, 2, 3 or 4. In other words, whacking the first hole did absolutely nothing to reduce the number of possible locations, and you were back to square one!

Suppose you instead started on Hole 2 and missed, in which case the mole could have been in Holes 1, 3 or 4. This time, on the next turn, the mole could have been in Holes 2, 3 or 4 — but not Hole 1, since it had to have moved from Hole 2, which you had just whacked. Next, if you whacked Hole 3 and missed, you could guarantee the mole was in either Hole 2 or 4. Whacking Hole 3 a second time and missing meant the mole now had to be in Hole 1. One last whack of Hole 2 ensured you got the mole.

At the end of the day, the best you could do with four holes was four whacks. The sequences 2-3-3-2 and 3-2-2-3 both guaranteed the job got done with minimal effort.

But what about 100 holes? Before we come back to that, let’s take one more look at the case of four holes, this time keeping in mind the initial parity of the mole’s position — as noted by solvers like Michael Coffey and Madeline Argent, and as suggested in the video that inspired this Classic. Every move, the mole either switched from an even-numbered hole to an odd-numbered hole, or vice versa. So every move, its parity flipped.

Suppose the mole had started in one of the even holes (i.e., 2 or 4). Checking Hole 2 and then Hole 3 guaranteed that you’d whack the mole. But had the mole started in an odd hole (i.e., 1 or 3), then after you checked Holes 2 and 3 it would still be in one of the odd holes. In this case, flipping your order from here on out — checking Hole 3 and then 2 — guaranteed that you’d whack the mole.

This exact line of reasoning extended to any number of holes, N. First, you could rule out all the even starting points by working your way up from 2 to N−1. Then, you could rule out all the odd starting points by returning from N−1 to 2. (This strategy even worked when N was odd.) So for N holes, you needed 2·(N−2) turns to guarantee that you whacked the mole. When N was 100, the answer was 196.

If you’re still not convinced, here’s an animation showing 100 moles that initially occupy all 100 holes. Each mole “reproduces,” splitting into two moles in adjacent holes, unless it’s at one of the end points, in which case it simply moves left or right. Occupied holes are red, while empty holes are white. Meanwhile, the hole you are currently whacking is blue.

Animation showing 100 holes all initially occupied. You whack from Holes 2 to 99, at which point alternating holes are occupied. As you work your way back from 99 down to 2, only holes on the left are occupied, until they are all empty at the end.

Sure enough, if you whacked your way up and down the row, you accounted for every possible initial position of the mole and every possible sequence of its moves.

Now if only we could handle tribbles with such efficiency.

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 [email protected].

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