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 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.
From Michael Fienberg comes a tower of not so much terror:
My condo complex has a single elevator that serves four stories: the garage (G), the first floor (1), the second floor (2) and the third floor (3). Unfortunately, the elevator is malfunctioning and stopping at every single floor, no matter what. The elevator always goes G, 1, 2, 3, 2, 1, G, 1, 2, etc.
I want to board the elevator on a random floor (with all four floors being equally likely). As I round the corner to approach the elevator, I hear that its doors have closed, but I have no further information about which floor it’s on or whether the elevator is going up or down. The doors might have just closed on my floor, for all I know.
On average, how many stops will the elevator make until it opens on my floor (including the stop on your floor)? For example, if I am waiting on the second floor, and I heard the doors closing on the garage level, then the elevator would open on my floor in two stops.
Extra credit: Instead of four floors, suppose my condo had N floors. On average, how many stops will the elevator make until it opens on my floor?
The solution to this Riddler Express can be found in the following column.
You are the coach at Riddler Fencing Academy, where your three students are squaring off against a neighboring squad. Each of your students has a different probability of winning any given point in a match. The strongest fencer has a 75 percent chance of winning each point. The weakest has only a 25 percent chance of winning each point. The remaining fencer has a 50 percent probability of winning each point.
The match will be a relay. First, one of your students will face off against an opponent. As soon as one of them reaches a score of 15, they are both swapped out. Then, a different student of yours faces a different opponent, continuing from wherever the score left off. When one team reaches 30 (not necessarily from the same team that first reached 15), both fencers are swapped out. The remaining two fencers continue the relay until one team reaches 45 points.
As the coach, you can choose the order in which your three students occupy the three positions in the relay: going first, second or third. How will you order them? And then what will be your team’s chances of winning the relay?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to ÑÑâÐ Steve Schaefer ÑÑâÐ of Carlsbad, California, winner of last week’s Riddler Express.
Last week was Hanukkah, which meant it was time for some Menorah Math!
I had a most peculiar menorah. Like most menorahs, it had nine total candles — a central candle, called the shamash, four to the left of the shamash and another four to the right. But unlike most menorahs, the eight candles on either side of the shamash were numbered. The two candles adjacent to the shamash were both “1,” the next two candles out from the shamash were “2,” the next pair were “3,” and the outermost pair were “4.”
The shamash was always lit. How many ways were there to light the remaining eight candles so that sums on either side of the menorah were “balanced”? For example, one such way was to light candles 1 and 4 on one side and candles 2 and 3 on the other side. In this case, the sums on both sides were 5, so the menorah was balanced.
Solver Madeline Argent approached this puzzle by first noting that the sum on either side could be any integer from 0 (when no candles were lit) to 10 (when all four candles were lit). However, many sums were achievable in more than one way, such as the aforementioned 5, which was both 1+4 and 2+3.
Here were all the possible sums, as well as how many ways they could be achieved:
- There was one way to have a sum of 0 (no candles lit).
- There was one way to have a sum of 1 (just candle 1 was lit).
- There was one way to have a sum of 2 (just candle 2).
- There were two ways to have a sum of 3 (just candle 3, or candles 1 and 2).
- There were two ways to have a sum of 4 (just candle 4, or candles 1 and 3).
- There were two ways to have a sum of 5 (candles 1 and 4, or candles 2 and 3).
- There were two ways to have a sum of 6 (candles 2 and 4, or candles 1, 2 and 3).
- There were two ways to have a sum of 7 (candles 3 and 4, or candles 1, 2 and 4).
- There was one way to have a sum of 8 (candles 1, 3 and 4).
- There was one way to have a sum of 9 (candles 2, 3 and 4).
- There was one way to have a sum of 10 (candles 1, 2, 3 and 4).
When the sums on either side were 0, 1, 2, 8, 9, or 10, there was only one way to balance the menorah, since each of these sums were only possible one way. But when the sums on either side were 3, 4, 5, 6 or 7, there were four ways to balance the menorah. For example, if the sums were 5, then both sides of the menorah could have been 1+4, both sides could have been 2+3, the left side could have been 1+4 and the right side 2+3, or the left side could have been 2+3 and the right side 1+4.
In all, there were six sums with one balanced lighting, and another five sums with four balanced lightings. That meant there were 26 balanced ways to light the menorah.
Perhaps Hanukkah should be 26 nights, instead of eight. I, for one, would not complain.
By the way, if you didn’t count the case where both sums were 0, which meant your answer was 25, I still counted that as correct. Who wants an empty menorah? (Looking at you, students at The Hewitt School!)
Solution to last week’s Riddler Classic
Congratulations to ÑÑâÐ Jake Gacuan ÑÑâÐ of Manila, Philippines, winner of last week’s Riddler Classic.
Last week, official ice master and deliverer Kristoff and his trusty pal Sven were preparing ice cubes that would be sent to the hot springs, where Kristoff’s extended family lived.
They were studying a cube whose side lengths measured 1 meter, which meant its surface-area-to-volume ratio is 6. (Kristoff didn’t particularly care about the units here but knew they were inverse meters.) Sven was concerned that much of the ice would melt before they reached the hot springs, so he suggested that Kristoff cut the ice along a single plane to minimize the surface-area-to-volume ratio of the larger resulting piece.
Along which plane should Kristoff have cut the ice? And what was the resulting surface-area-to-volume ratio?
First off, several solvers noted that any cut, regardless of the resulting surface-area-to-volume ratio, would have made the ice melt faster. To that I say that Sven is just a reindeer, and his mathematical prowess exceeds his knowledge of thermodynamics.
Back to the ice — for any given surface area (or volume), the 3D solid with the minimal surface-area-to-volume ratio is the sphere. And so one reasonable approach was to make the cube of ice a little more “spherical” (somehow).
To convince yourself this was even possible, you could have cut off a small corner of the cube along a plane that intersected three edges a distance x from one vertex, as shown below.
In this case, you were slicing off a tetrahedron with volume x3/6. You were also removing a surface composed of three right triangles with a combined area of 3x2/2. At the same time, you were creating a new equilateral surface with side length x√2, and hence area (x2√3)/2. Putting this all together, the surface-area-to-volume ratio was (6−(3−√3)x2/2)/(1−x3/6). For small values of x, the denominator was very close to 1, while the numerator was slightly less than 6. In other words, it was indeed possible for the ratio to be less than 6!
Minimizing the aforementioned rational function gave you an optimal ratio of roughly 5.962 when x was approximately 0.4254. And that’s the answer that several readers submitted. But was it possible for Kristoff to do even better?
Another way to make the cube slightly more “spherical” was to cut along a plane that was parallel to an edge:
If the distance from the lopped-off corners to where the plane intersected the edges was again x, the volume removed this time was x2/2, while the surface area removed was x2+2x. Meanwhile, the newly created surface area was x√2. This time around, the resulting ratio was (6−(2−√2)x−x2)/(1−x2/2). Minimizing this rational function gave you a ratio of approximately 5.957 when x was about 0.1481. As it turned out, cutting parallel to an edge was slightly better than lopping off a corner.
This week’s winner, Jake, extended the puzzle by looking at what would happen if Kristoff made multiple cuts, slicing off multiple corners or edges. According to Jake, with eight cuts, Kristoff could achieve a surface-area-to-volume ratio as low as 5.66. And if you want to see how solver Tyler Barron got this week’s answer in real time, check out his work on Twitch!
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 firstname.lastname@example.org.