Welcome to The Joker. 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 Joker Express for those of you who want something bite-size and the Joker 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.
This week’s Express is an oldie but a goodie:
Multiplying polynomials can be challenging work. But every once in a while, there’s an expansion that is particularly elegant. One of my favorites is the following expression:
The ellipsis indicates that every letter is subtracted at some point. If you multiply all these binomials together, can you write the expansion in simplest form?2
The solution to this Joker Express can be found in the following column.
To play the game of “Zactoll” (named for yours truly, of course), you start with the number 1 and then try to reach a target number through a series of steps. For each step, you can always choose to double the number you currently have. However, if the number happens to be one more than an odd multiple of 3, you can choose to “reduce” — that is, subtract 1 and then divide by 3.
For example, to reach a target number of 5, you could double (1 → 2), double (2 → 4), double (4 → 8), double (8 → 16) and reduce (16 → 5). And to get a target number of 6, you could double (1 → 2), double (2 → 4), double (4 → 8), double (8 → 16), reduce (16 → 5), double (5 → 10), reduce (10 → 3) and double (3 → 6).
What is the smallest target whole number you cannot reach in the game of Zactoll?
The solution to this Joker Classic can be found in the following column.
Solution to last week’s Joker Express
Congratulations to 👏 Tyler Cochran 👏 of Acworth, Georgia, winner of last week’s Joker Express.
Last week, you had a case with 11 golden globes, weighing 1 kilogram, 2 kg and so on, up to 11 kg. And you knew exactly which globe was which.
You had arranged to sell one of the globes to a queen. She had heard tales of these orbs, and knew for a fact that they had masses from 1 kg to 11 kg. However, she didn’t know which was which and would not simply take your word for it. She agreed to purchase a globe as long as you could demonstrate its weight.
The queen had a balancing scale with two plates, one on each side. It showed whether the total weight on either plate was equal or, if not, which side was heavier. The queen could clearly see which globes you placed on which plate. However, if at any point the mass on either side exceeded 11 kg, the scale would break and the queen would refuse to buy from you.
The queen was in a hurry for her next appointment, so time was limited. What was the fewest number of weighings by which you could prove the weight of at least one globe? Which globe was it?
A good way to begin was to think from the queen’s perspective. Which placements of the globes would give her as much information about them as possible? As noted by reader Rose Feuer, there were only two ways to place four globes on one side of the scale without breaking it: 1 + 2 + 3 + 4 (with a total weight of 10 kg) and 1 + 2 + 3 + 5 (with a total weight of 11 kg). So the moment you put four weights on one side and nothing breaks, the queen can immediately deduce that three of those weights are 1, 2 and 3, and that the final weight is 4 or 5 kg. However, she doesn’t yet know which is which.
Suppose you had put the 1, 2, 3 and 4 kg globes on one side, and the 10 kg globe on the other, resulting in a balanced scale. At this point, the queen knew the globe on the plate by itself had to be either 10 or 11 kg. Next, you could remove the 1, 2, 3 and 4 kg globes and replace them with the 11 kg one, which was heavier than the 10 kg globe. At this point, the queen would realize the previous globe in question was 10 kg, since there existed a heavier globe. And so, after just two weighings, she knew which globe was 10 and which was 11 kg. But was it possible to do even better?
Indeed, there was a way to demonstrate one of the weights in a single attempt. If you had put the 1, 2, 3 and 4 kg globes on one side, and the 11 kg globe on the other, the scale would have been imbalanced in favor of the 11 kg globe. That was the only way to have a single globe outweigh four others, which meant after just one weighing you could sell the 11 kg globe.
It was only fitting that the queen purchased the most massive of the globes.
Solution to last week’s Joker Classic
Congratulations to 👏 Simon Birenbaum 👏 of Baltimore, Maryland, winner of last week’s Joker Classic.
Last week, the astronomers of Planet Xiddler had developed a new technology for measuring the radius of a planet by analyzing its cross sections. They launched a satellite to study a newly discovered, spherical planet. The satellite sent back data about three parallel, equally spaced circular cross sections which had radii A, B and C megameters, with 0 < A < B < C. Based on these values, the scientists calculated that the radius of the planet was R megameters. To their astonishment, they found that A, B, C and R were all whole numbers!
What was the smallest possible radius of the newly discovered planet?
The responses to this puzzle were all over the map. Many readers thought the answer was 65 (i.e., that R was 65 megameters). That’s because 65 happened to be the length of the hypotenuse for three distinct Pythagorean triples with legs that formed an arithmetic sequence. These three triangles were 25-60-65, 33-56-65 and 39-52-65. Those middle numbers — 60, 56, and 52 — represented the distances from the three planes to the center of the spherical planet, and they were evenly spaced. The other lengths — 25, 33 and 39, shown below — represented values for A, B and C, the radii of those circular cross sections.
While this approach certainly generated integer values for A, B, C and R, it didn’t result in the smallest possible radius. These Pythagorean triples had the property that the distances between the three planes were also integers — in this case, 4 megameters. However, having integer distances between planes was not a requirement of the problem!
Another way to proceed was to find an expression for R in terms of A, B and C. Suppose the plane containing the circle with radius B was a distance k from the planet’s center, while the other two planes were a distance h from that plane. By applying the Pythagorean theorem to each circle, you got A2 + (k+h)2 = R2, B2 + k2 = R2 and C2 + (k−h)2 = R2. From there, you could add the first and third equations and subtract the second equation twice, giving you an expression for h: h2 = (2B2−A2−C2)/2. Next, subtracting the first equation from the third gave you an expression for k: k = (C2−A2)/(4h). Finally, plugging these back into the second equation gave you the equation for R: R2 = B2 + 1/8·(C2−A2)2/(2B2−A2−C2).
From there, most solvers wrote some code to try out integer values of A, B and C to find when R was also an integer, with apologies to anyone looking here for a more elegant method from the equations alone. As it turned out, the smallest such solution with A < B < C ≤ R (the requirement given in the puzzle) occurred when A was 2, B was 7 and C was 8. With these three values, R also turned out to be 8 megameters. In this case, the distance between the planes was √15.
Solver Rohan Lewis took this puzzle a step further, looking at cases where A, B, and C were all strictly less than R, as well as when the three cross sections didn’t all lie on the same half of the planet.
By the way, Earth’s radius is a little less than 6.4 megameters. With a radius of 8 megameters, the Xiddlerians might have just discovered a habitable planet.
Want more puzzles?
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 Joker,” and it’s in stores now!
Want to submit a puzzle?
Email Zach Wissner-Gross at email@example.com.