Can you outsmart the tax collector?

Welcome to the Riddler. Each week, I come up with problems related to things that are dear to us here: math, logic, and probability. Two puzzles are featured each week: the Riddler Express for those of you who want something small and the Riddler Classic for those of you who like the movement of slow puzzles. Submit a correct answer for either, and you might get a shoutout in the next column. Please wait until Monday to share your answers publicly! If you need a clue or if your favorite puzzle is gathering dust in your attic, find me on twitter.

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

Riddler Express

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

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

The problem? This puzzle has of them possible solutions – i.e. two distinct sets of letter-digit assignments. Can you find both solutions?

The solution to this Riddler Express is in the next column.

classic riddler

This week’s classic is an expansion of a puzzle originally created by Dan Finkel (and was relayed to me by Fawn Nguyen):

In the “tax collector” game, there are paychecks with different whole dollar amounts, from 1 to NOT. You choose a salary, one at a time. For any check you choose, the tax collector immediately takes all remaining checks (i.e., not already picked up by you or the tax collector) whose dollar values ​​are factors of the one you you chose. For example, if you choose the check for $10, the tax collector will immediately take the checks for $1, $2, and $5 – if one is available. Above all, the tax collector must get something for each salary you choose. So if the $10 check is available, but the $1, $2, and $5 checks are not, you cannot accept 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 objective was to earn more money than the tax collector when NOT was 12 (or 24 or 48). When NOT was 12, you could earn $50 of the tax collector’s $28, which means you earned about 64% of the total. When NOT was 24, you could earn about 61% of the total, and when NOT was 48, you could earn about 62%.

For this puzzle, not only do you want to get more money than the tax collector, but you also want to earn the largest possible fraction of the available money. What value of NOT (greater than 1) would you choose, so that you could win most of the available money?

The solution to this Riddler Classic is in the next column.

Solution to last week’s Riddler Express

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

Last week you were assigned to paint 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 – a central blue triangle surrounded by three white triangles.

This 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.  In profile, we notice that each of the three side faces carries a blue kite.

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

It first helped realize that the fraction of the total area that was blue was the same for each face, whether you looked at it from above or from the front. (For a good explanation of why, watch the recent video from 3Blue1Brown.) In other words, a quarter of each of the three faces were painted blue, and the kites met the side edges of the tetrahedron three-quarters of the way . at the top.

The answer did not depend on the overall scale of the tetrahedron, so let’s say it had side length 4, as shown in the diagram below. This meant that the top two edges of the kite had a length of 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, i.e. 2√3. Again, three quarters of this height was below half the width of the kite, for a total length of 1.5√3.

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

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

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

Solution to last week’s Riddler Classic

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

Last week, you were trying to hit a mole. Every second, the mole poked its head out of one of the 100 holes arranged in a line. You didn’t know where the mole started, but you do know that it was always moving from one hole to an adjacent hole every second. For example, if he is out of the 47th hole, one second later he is out of the 46th hole or the 48th hole. If he exited the first hole, he was guaranteed to exit the second hole next; similarly, if he exited the 100th hole, then he was guaranteed to exit the next 99th hole.

Of course, there was a catch – the mole was camouflaged and you had no idea where it was at all times until you actually hit it. Every second you can drill a hole of your choice. Many shot sequences didn’t guarantee you’d end up with the mole (e.g. always hitting the first hole), but some did.

What was the shortest streak of this guy that guaranteed you could hit the mole no matter where it started or how it moved?

Before dealing with 100 holes, consider a simpler case: four holes labeled 1, 2, 3, and 4. Suppose you first hit 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 d In other words, hitting the first hole did absolutely nothing to reduce the number of possible locations, and you were back to square one!

Suppose you instead started at hole 2 and missed, in which case the mole could have been in holes 1, 3 or 4. This time in the next round the mole could have been in holes 2, 3 or 4 – but not hole 1, since it must have moved from hole 2, which you had just hit. Then, if you hit hole 3 and miss, you can guarantee the mole was in hole 2 or 4. Hitting hole 3 a second time and missing meant the mole now had to be in hole 1. One last shot of hole 2 assured you got the mole.

At the end of the day, the most you could do with four holes was four shots. The 2-3-3-2 and 3-2-2-3 sequences both ensured the job was done with minimal effort.

But what about 100 holes? Before we go back to that, let’s look at the four-hole case one more time, 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. With each move, the mole moved from an even hole to an odd hole, or vice versa. Thus, with each movement, its parity was reversed.

Suppose the mole started in one of the even numbered holes (i.e. 2 or 4). Checking hole 2 then hole 3 guaranteed you would hit the mole. But if the mole had started in an odd hole (i.e. 1 or 3), then after checking holes 2 and 3, it would have always be in one of the odd holes. In this case, returning your order from now on – checking hole 3 then 2 – guaranteed you’d hit the mole.

This exact reasoning extended to any number of holes, NOT. First, you can exclude all even starting points by going up from 2 to NOT−1. Then you can eliminate all odd starting points by returning from NOT−1 to 2. (This strategy even worked when NOT was strange.) So for NOT holes, you needed 2 (NOT-2) spins to guarantee you hit the mole. When NOT was 100, the answer was 196.

If you’re still not convinced, here’s an animation showing 100 moles initially occupying the 100 holes. Each mole “reproduces”, splitting into two moles in adjacent holes, unless it is at one of the endpoints, in which case it simply moves left or right. Occupied holes are red, while empty holes are white. During this time, the hole you are digging is blue.

Animation showing 100 holes initially all occupied.  You hit holes 2 through 99, at which point alternate holes are occupied.  As you go back from 99 to 2, only the left holes are occupied, until they are all empty at the end.

Effectively, if you worked your way up and down the row, you were counting all possible initial positions of the mole and all possible sequences of its movements.

Now if only we could handle tribbles with such efficiency.

Want more puzzles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles in this column and some never-before-seen puzzles. It’s called “The Riddler”, and it’s in stores now!

Want to submit a puzzle?

Email Zach Wissner-Gross at [email protected]

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Esther L. Steinbach