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 next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
Riddler Express
This week, Eli Luberoff presents a puzzle in which artistry meets geometry. While at the San Francisco Museum of Modern Art, Eli snapped the following picture from an exhibit by Sol LeWitt:
What’s the volume of the darkened solid?
(Assume the outer cube is a unit cube, and that the faces of the solid meet the edges of the unit cube halfway and a fourth of the way along its edges. In other words, the drawing is to scale!)
Riddler Classic
I want to carve the perfect eye for my pumpkin this Halloween, but I can’t seem to make it the right size. Since symmetry is the key to beauty, the triangular eye should be equilateral and equiangular, which is easier said than done on the surface of a spherical pumpkin!
To be a proper triangle, the three corners should be connected by arcs that are “straight,” meaning they go directly from one corner to the next via the shortest possible path. (Think about air travel: When you flying from the West Coast of the U.S. to Europe, you’ll fly north of the Arctic Circle along the way, since that’s the shortest path over the Earth’s curved surface.)
Anyway, the triangular eye that I made is way too big. Its sides all meet at right angles, and the resulting eye takes up a whole eighth of the pumpkin’s surface, as seen in the animation below.
Instead, I want an eye that’s precisely half that size, or one-sixteenth of the pumpkin’s surface. For such an ideal pumpkin eye, at what angle should each of the sides meet?
Solution to last week’s Riddler Express
Congratulations to ÑÑâÐJames Sears ÑÑâÐ of Saint John, New Brunswick, Canada, winner of last week’s Riddler Express.
Last week, you had to navigate your way through the number maze below. The number in each box told you how many spaces up, down, left or right you must move to get to the next box. You started at the yellow 6 in the bottom left corner, and you were challenged to make your way to the asterisk.
James found his solution by working backwards. First, he tried to find a square from which you could get to the asterisk. Such a square would have to be in the same row or column as the asterisk. The only square that works is the 2 that’s two squares left of the asterisk — no other square in the asterisk’s row or column has a number that equals how many squares away from the asterisk it is. Next, from which squares could you get to that 2? The only such square is a 6 that’s six rows up. Continuing in this fashion, James found a nine-step path that works:
Much of Riddler Nation took this puzzle one step further, trying to find the shortest path to the asterisk. David Schwab of Ottawa, Ontario, Canada (man, the Canadians crushed it this week!) performed a breadth-first search, starting with the 6 in the lower left and finding all the squares you could move to, and labeled them as reachable within one step. Then, he found all the squares you could get to from those squares and labeled them as reachable within two steps. He continued this process until all the squares were reached, resulting in the following guide through the maze:
The number in the top left corner of each square indicates how many moves are required to reach that square, starting from the 6 in the lower left corner. It turns out that the asterisk can be reached in eight moves. Meanwhile, that 7 in the top right is the toughest to reach, requiring a whopping 16 moves. Who knew?
Solution to last week’s Riddler Classic
Congratulations to ÑÑâÐKyle Tripp ÑÑâÐ of Concord, California, winner of last week’s Riddler Classic.
Last week you converted dollars into the currency of Riddler Nation: the Dio, worth $538, and the Phantus, worth $19. For example, if you wanted to convert $614, you’d get one Dio and four Phanti, since 614 = 1×538 + 4×19. But if you tried to exchange one dollar more (i.e., $615), then alas, no combination of Dios and Phanti would work. What was the largest number of dollars that could not be converted into Dios and Phanti?
The name of the currency was a clue. Diophantus was a Greek mathematician and the namesake of Diophantine equations, which are polynomial equations that only seek integer solutions. For this particular riddle, the Diophantine equation of interest was 538x + 19y = z, where x represents the number of Dios, y is the number of Phanti, and z is the number of dollars.
Solver Guy Moore solved the problem by looking at increments of 538. For dollar amounts less than 538, only multiples of 19 are exchangeable, all the way up to 19×28 = 532.
Beyond 538, you have more possibilities because we’re working with a second denomination. You can now exchange dollar amounts that are multiples of 19 (in other words, that have a remainder of 0 when divided by 19), as well as dollar amounts that can be exchanged for exactly one Dio and any number of Phanti, such as 538, 557 (538 + 19), 576 (538 + 19 + 19) and so on. You’ll notice that all those numbers share something in common: When divided by 19, they have a remainder of six. That’s because when the number 538 is divided by 19, it leaves a remainder of six.
All this talk of remainders might seem superfluous, but they’re important for understanding how to get to a solution.
When you reach 1,076 (538×2), you can exchange dollar amounts that are multiples of 19, dollar amounts that have a remainder of six when divided by 19 and dollar amounts that have a remainder of 12 when divided by 19. (That’s because when the number 1,076 is divided by 19, it leaves a remainder of 12.)
If you keep going, numbers with different remainders (when divided by 19) become exchangeable every time you hit another multiple of 538. All the numbers become exchangeable when, at last, you reach the 18th (i.e., one less than 19) multiple of 538 — 9,684 — which happens to have a remainder of 13 when divided by 19. Beyond 9,684, every single dollar amount — no matter its remainder when divided by 19 — is exchangeable. For example, what about a large number like 100,000? That equals 538×10 + 4,980×19. That’s a lot of Phanti!
But we still have to find an answer! Let’s return to our discovery of 9,684, which was the first multiple of 538 with a remainder of 13 when divided by 19. (The 13 isn’t very important here — it just happens to be the last of the 19 possible remainders to become exchangeable.) That means numbers less than 9,684 with the same remainder of 13 are not exchangeable. The largest of these is 9,684 – 19 = 9,665, the correct answer.
Many solvers were also quick to point out that this riddle was essentially asking for the Frobenius number, which is the largest value that a set of numbers cannot generate when multiplied by whole numbers and then added together. The Frobenius number for two numbers whose greatest common factor is one can be found by subtracting one from each number, multiplying those values and then subtracting one from that product. In this case, that’s (538 – 1)(19 – 1) – 1, which indeed equals 9,665.
For extra credit, you were asked how your answer would change upon the inclusion of a third currency, worth $101. Guy observed that the Dio would then be equivalent to 23 Phanti plus this new $101 denomination. That means that any time you exchange dollars for a Dio, you could have similarly exchanged it for Phanti and $101s. In other words, the Dio just became obsolete in our calculations! Therefore, the solution to the extra credit was simply the Frobenius number for 101 and 19, which is 1,799.
So when you travel to Riddler Nation, be sure to try to exchange exactly $9,665. They might throw you out of the country, but they’d just as likely applaud your mastery of number theory.
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.