Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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
It’s currently snowing outside my window, in what I believe is the 37th or 38th nor’easter to hit so far this month. But my mood is buoyed by the coming of summer, which this year will bring an extra treat: the men’s World Cup!
In the World Cup’s group stage, each group of four teams plays a round robin, with every team playing the other teams once each. A win is worth 3 points, a draw 1 point, and a loss 0 points. The points are tallied, and the top two teams in each group advance to the knockout stage. For a given group — let’s pick Group A, which includes Russia, Saudi Arabia, Egypt and Uruguay — how many different final standings, including point totals, are possible for the group stage?
Riddler Classic
From Joseph Converse, a puzzle of digital manipulation:
Imagine taking a number and moving its last digit to the front. For example, 1,234 would become 4,123. What is the smallest positive integer such that when you do this, the result is exactly double the original number? (For bonus points, solve this one without a computer.)
Solution to last week’s Riddler Express
Congratulations to 👏 Monica Cataldo 👏 of Derry, New Hampshire, winner of last week’s Riddler Express!
Last week you were presented with the following shapely arithmetic equation:
So what does a hexagon divided by a nonagon equal, anyway? It equals 6/7.
Or that was the solution intended by puzzle submitter Derik Moore and me. The rationale lies in the shapes’ internal angles. One angle in an equilateral triangle equals 60 degrees, and one angle in a regular hexagon equals 120 degrees, and 60/120 = 1/2. Similarly, one angle of a square is 90 degrees, and one angle of an octagon is 135 degrees, and 90/135 = 2/3. And, finally one angle of a hexagon is 120 degrees, and one angle of a nonagon is 140 degrees, and 120/140 = 6/7, our answer.
Other solvers found more exotic explanations for their solutions. Sion Verschraege, for example, inscribed each shape into a unit circle and took the ratios of their areas. He wound up with a solution of \(\frac{1}{\sin (2\pi/9) \cdot \sqrt{3}}\). Another solver wrote that it was “easy peasy to fit a curve to a two-data-point trendline with whatever the heck function I want, because you didn’t stipulate.” Riddler Nation’s imagination is as great as its calculations.
Solution to last week’s Riddler Classic
Congratulations to 👏 Jason Shaw 👏 of Topeka, Kansas, winner of last week’s Riddler Classic!
Last week I posed a seemingly simple question: How many decimal numbers are there whose value equals the average of their digits? The digits of the number 2.3, for example, are 2 and 3, and their average is 2.5 — so that doesn’t quite work. But there are at least a few that do.
In fact, it appears there are an infinite number of such numbers. (I’ll call them Bollier numbers, after this puzzle’s submitter, Sam Bollier.)
It’s not too hard to find your first Bollier number. Single-digit whole numbers are obvious, if trivial, solutions. The average of the digits of the number 7, for example, is just 7. The number 4.5 works well, too — its digits are 4 and 5, and their average is 4.5. Bingo. The number 3.83333 does, too. The average of its digits is (3+8+3+3+3+3)/6 = 3.833333…, which we are allowed to truncate according to the rules of the riddle. There are more complicated numbers too. 4.428571 works, as does 4.571429. And so does 4.5294117647058824.
But when can we stop looking for Bollier numbers? Solver Jason Weisman used a computer program to find hundreds of Bollier numbers with less than 1,000 digits, and he reported that their density on the number line remained more or less consistent as he kept searching. Stephen Cappella found a really long one: the first 11,617 digits of the decimal expansion of 52,277/11,617. Stephen also graphed the averages of the digits for the first 8,000 or so Bollier numbers he found:
(The cluster of averages around 4.5 is interesting, but mysterious. If Riddler Nation has any mathematical insights into this cool pattern, please let me know!)
Similarly, solver Lucas Jacobson plotted the number of solutions as the number of digits grew. The number of Bollier numbers appears to increase without bound.
But can we prove that there are an infinite number mathematically, and not just illustrate it empirically?
Solver Eric Eason and others argued that the proof lies in prime numbers. Eric emailed me a recipe for constructing any amount of arbitrarily large Bollier numbers:
- Pick a prime number, N.
- Compute the number 4.5 – 0.5*(1/N).
- Truncate this number to N digits.
This will lead to a Bollier number, Eric wrote, if the prime has the following property: The repeating part of the decimal representation, also called a repetend, of 1/N must have an even number of digits. For example, 7 works, because it has a repetend of length six: 1/7 = 0.142857 142857 142857… Same for 11, which has a repetend of length two: 1/11 = 0.09 09 09… Conversely, 3 and 31 do not work because they have repetends of length one and 15. About 78 percent of the first 10,000 primes have the desired property, so it seems likely that there are an infinite number of such primes.
There’s even more to this intricate proof idea, which also relies on something called Midy’s theorem. But suffice it to say that you will likely never run out of Bollier numbers.
Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.