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.
It’s time for a random number duel! You and I will both use random number generators, which should give you random real numbers between 0 and 1. Whoever’s number is greater wins the duel!
There’s just one problem. I’ve hacked your random number generator. Instead of giving you a random number between 0 and 1, it gives you a random number between 0.1 and 0.8.
What are your chances of winning the duel?
The solution to this Riddler Express can be found in the following column.
From Gary Yane comes a question that launched a million flips:
I have in my possession 1 million fair coins. Before you ask, these are not legal tender. Among these, I want to find the “luckiest” coin.
I first flip all 1 million coins simultaneously (I’m great at multitasking like that), discarding any coins that come up tails. I flip all the coins that come up heads a second time, and I again discard any of these coins that come up tails. I repeat this process, over and over again. If at any point I am left with one coin, I declare that to be the “luckiest” coin.
But getting to one coin is no sure thing. For example, I might find myself with two coins, flip both of them and have both come up tails. Then I would have zero coins, never having had exactly one coin.
What is the probability that I will — at some point — have exactly one “luckiest” coin?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Grant Bronsdon 👏 of Seattle, Washington, winner of last week’s Riddler Express.
Last week, you looked at trapezoidal numbers, which are whole numbers that can be written as the sum of two or more consecutive positive integers. For example, 9 is a trapezoidal number because it can be written as 2+3+4. In fact, 9 is doubly trapezoidal because it can also be written as 4+5.
Your challenge was to find the smallest number that is triply trapezoidal, meaning it is trapezoidal in three different ways.
A direct approach was to try out some different numbers. Eventually, you would have found that 15 was the smallest triply trapezoidal number, as it equals 1+2+3+4+5, 4+5+6 and 7+8.
That wasn’t too bad. But things got trickier with the extra credit, which asked for the smallest quadruply, quintuply and sextuply trapezoidal numbers.
Before we answer that, it’s worth exploring some of the patterns that trapezoidal numbers exhibit. For example, every multiple of 3 (greater than 3) can be written as the sum of three consecutive positive integers, since 3N = (N−1)+N+(N+1). Similarly, every multiple of 5 (greater than 5) can be written as the sum of five consecutive positive integers. The same goes for 7, 9, 11 and any other odd number. That explains why 15 was a good candidate for an answer, since it was a multiple of two small odd numbers, 3 and 5.
Now what about even numbers? Multiples of 2 can’t be written as the sum of two consecutive integers. However, every odd number — meaning numbers that are congruent to 1 (mod 2) — can be. Numbers that are 2 (mod 4) can be written as the sum of four consecutive integers, and numbers that are congruent to 3 (mod 6) can be written as the sum of six consecutive integers. In general, half-multiples of even numbers can be written as the sum of that many consecutive integers.
In the case of 15, it was congruent to 1 (mod 2). And while 15 was also congruent to 3 (mod 6), it wasn’t large enough to be written as six consecutive positive integers. (Instead, it was 0+1+2+3+4+5.)
At this point, you were on the hunt for multiples of odd numbers and half-multiples of even numbers.
The next number worthy of inspection after 15 was 45, which was a multiple of 3, 5 and 9. It was also odd, meaning it was a half-multiple of 2, as well as a half-multiple of 6. Technically, 45 was quintuply trapezoidal! To find a number that was trapezoidal in exactly four ways, you needed to go all the way up to 81, which was a multiple of 3, 9 and 27, while also being a half-multiple of 2.
We just said 45 was quintuply trapezoidal. But was it also, technically, quadruply trapezoidal, since it was trapezoidal in (at least) four ways? Riddler Nation was split on this, so I take no issue with accepting inclusive and exclusive definitions — at least for this riddle. (Parallelograms are still trapezoids, people!)
Team Inclusive found the first quadruply, quintuply and sextuply trapezoidal numbers were 45, 45 and 105, respectively. Meanwhile, Team Exclusive found they were 81, 45 and 729.
The trapezoidal nature of the numbers from 1 to 81 was captured in a single chart by solver Jenny Mitchell, as shown below. Each number was fractured into a number of parts indicating how many trapezoids it could form.
By the way, Team Inclusive had its own preexisting OEIS sequence to back them up. Just saying.
Solution to last week’s Riddler Classic
Congratulations to 👏 Seth DeLand 👏 of Ida, Michigan, winner of last week’s Riddler Classic.
Last week, you considered the eight vertices (or corners) of a cube. Suppose I had assigned a prime number to each vertex. A “face sum” was the value I got when I added up all four prime numbers on one of the six faces.
Could you find eight primes and arrange them on a cube so that the six face sums were all equal?
Hats off to folks who assigned the same prime number to all eight vertices. You got a good chuckle at the expense of my imprecise wording. But anyway, this problem was much more interesting when you (rightfully) assumed that all eight primes had to be distinct.
Several solvers brute forced their way through this, turning to their computers to search for eight satisfactory prime numbers.
Meanwhile, solver Diarmuid Early was content to (cleverly) prove that infinitely many solutions existed. Let’s start there!
First off, you could place the numbers from 1 to 8 on the vertices of a cube so that all six face sums were 18:
Diarmuid observed that this numerical assignment worked not only for these eight numbers, but for any arithmetic sequence with eight terms. That was because multiplying all eight numbers by a constant and then adding another constant to all eight terms kept all six face sums equal. Now, the Green-Tao theorem states that prime numbers occur in arbitrarily long arithmetic sequences. It just so happens that the first such eight-term sequence is 199, 409, 619, 829, 1,039, 1,249, 1,459 and 1,669. Sure enough, this yields a solution in which each face sum is 3,766:
Of course, you weren’t expected to find this particular arrangement of primes. In particular, solver Laurent Lessard showed that the eight numbers didn’t have to form a single arithmetic progression, but rather that four of the numbers — placed at four corners forming a regular tetrahedron — were offset from the other four corners by a fixed amount, and that corresponding numbers that differed by that offset were placed on opposite corners.
There were infinitely many sets of eight primes that satisfied these conditions. The smallest face sum turned out to be 64, with the following arrangement:
Many other solutions were found by Riddler Nation, and infinitely many exist.
By the way, if you enjoyed this riddle and are looking for even more of a challenge, its author suggests you replace the cube with other platonic solids. Good luck!
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.