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Can You Turn Digits Into Numbers?

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 or send me an email.

Riddler Express

From Jeremy Dixon comes a “toasty” treat of a puzzle:

You have a toast rack with five slots, arranged in an array. Each slot has a slice of toasted bread, which you are removing one at a time. However, you are quite superstitious, and you know it’s bad luck to remove adjacent pieces of toast one after the other. (What? You’ve never heard that before? It’s totally a thing!)

How many different ways can you remove the slices of toast?

Extra credit: Instead of five slots, suppose you have a rack with six slots and six slices of toast. Now how many different ways can you remove the slices without ever removing adjacent pieces one after the other?

Submit your answer

Riddler Classic

The New York Times is beta-testing a new math game called Digits. In each level of the game, you are presented with six numbers and the four basic operations (addition, subtraction, multiplication and division). For example, the six numbers in the game below are 2, 3, 5, 14, 25 and 15.

Six numbers are shown: 2, 3, 5, 14, 25 and 15. Below them are five buttons: reset, add, subtract, multiply and divide.

With each step of the game, you first pick a number, then an operation and then another number. So if you picked 15, then ×, then 5, the 15 and 5 would disappear and be replaced by 75, as shown below:

Five numbers are shown: 2, 3, 75, 14 and 25. The 75 is highlighted. The same five buttons below are shown: reset, add, subtract, multiply and divide.

At this point, you could use the 75 similarly to how you use the other four numbers. The objective of the game is to use the numbers and operations to reach a specific target number. But let’s put that aside for now.

Instead, the question here is this: What is the greatest number of distinct values you can make using anywhere from one number by itself, all six numbers or anywhere in between? For the purposes of this puzzle, you can take your pick of starting numbers. Also, negative numbers and fractions are allowed — they can be starting numbers or values generated along the way.

Submit your answer

Solution to the last Riddler Express

Congratulations to Ñб─Ñб÷âб─б≤Ðб▐ Steve Schaefer Ñб─Ñб÷âб─б≤Ðб▐ of Carlsbad, California, winner of last week’s Riddler Express.

Last week, you considered the infinite points in the coordinate plane, and supposed that each point is one of three colors: red, green or blue. It turned out there had to be at least two points of the same color that were a distance 1 apart. But it was up to you to prove it, using just seven points in the plane.

If the points could only be one of two colors (e.g., red and blue) rather than three, you only needed three such points, arranged in an equilateral triangle with side length 1. All three vertices were a distance 1 from each other, and at least two of them had to be the same color, whether red or blue.

But with three colors, the puzzle was a good deal trickier. Several readers thought to arrange the seven points into a regular hexagon with side length 1 plus a point at its center. However, it was possible to color these seven points in such a way that any pair of points a distance 1 apart were different colors. One such coloring is shown below, with the alternating red and green vertices and a blue point at the center.

A regular hexagon is shown. The center point is labeled Blue. The vertex to the right is labeled Red. Clockwise around the perimeter from there the vertices are Green, Red, Green, Red and Green.

The solution was a graph known as the Moser spindle, illustrated below by solver Peter Exterkate.

Two rhombuses, which are themselves composed of equilateral triangles with a common edge. The rhombuses have a common vertex (at a 60 degree angle) that is colored red. The points in the middle of the rhombuses (that is, the end points of the common edges of the equilateral triangles) are colored blue and green. The vertices opposite from the common vertex are both colored red, and the rhombuses are rotated such that these two red points are a unit distance apart.

This “spindle” consisted of two rhombuses with side length 1, themselves composed of two equilateral triangles each. These rhombuses shared a common vertex, but they were rotated about this vertex in such a way that their opposite vertices were a distance 1 apart.

Whatever color you assigned to the common vertex, the four adjacent points had to be assigned the other two colors. That meant the two vertices opposite from the common vertex had to be the same color as the common vertex. Since these two vertices were themselves a distance 1 apart, you could conclude that the spindle needed at least four colors to avoid having points a distance 1 apart with the same color.

A fascinating extension of this puzzle, known as the Hadwiger-Nelson problem, asks for the minimum number of colors needed to color every point in the plane so that no points a distance 1 apart have the same color. You might intuitively think this number should be very large, but it turns out to be less than 8. The Moser spindle rules out 3, while 4 was also ruled out by several graphs between 2018 and 2020, containing hundreds or thousands of points.

The so-called “chromatic number” of the coordinate plane is therefore 5, 6 or 7 — and no one knows (yet) which of these three numbers is the right answer!

Solution to the last Riddler Classic

Congratulations to Ñб─Ñб÷âб─б≤Ðб▐Eric Snyder Ñб─Ñб÷âб─б≤Ðб▐ of Everett, Washington, winner of last week’s Riddler Classic.

Last week, you explored a novel proof of the Pythagorean Theorem by high school students Ne’Kiya Jackson and Calcea Johnson from St. Mary’s Academy in New Orleans, Louisiana.

Their proof applied the law of sines (which itself can be derived from equivalent expressions for a triangle’s area and has no dependency on the Pythagorean theorem, thereby avoiding any circular logic) to the construction below:

A right triangle is shown in red, with horizontal leg a, vertical leg b (with b greater than a) and hypotenuse c. The right angle is in the bottom right. A second, congruent right triangle is reflected across leg b.

The hypotenuse of the second right triangle is extended down and to the right. Another line extends down and to the right from the leftmost vertex of the first triangle, forming a right angle with that triangle's hypotenuse.

The two lines intersect in the bottom right. Prior to their intersection, the former has length q and the latter has length p.

Atop the figure were two reflected right triangles with legs a and b (with a b) and hypotenuse c. Below these triangles were what the students called a “waffle cone” shape, formed between the extensions of one of the top triangle’s hypotenuse and a line that was perpendicular to the other hypotenuse.

In their proof, they computed distances p and q, where p extended from the leftmost vertex of the two triangles to the intersection of the lines, and q extended from the topmost vertex of the two triangles to the same intersection.

Your challenge was to determine expressions for p and q in terms of a, b and c. However, in doing so, you couldn’t use the Pythagorean theorem in any of its forms (e.g., the so-called “distance formula,” etc.). After all, solving for p and q was a key step toward proving the Pythagorean theorem!

Jackson and Johnson subdivided the waffle into infinitely many similar triangles and then used geometric series to compute expressions for p and q. Solver Jim Jacobson took a similar approach, generating the following diagrams to solve for p and q:

The waffle, now subdivided into infinitely many similar triangles. Along the side with length p, these hypotenuses have length 2ca/b, 2ca^3/b^3, 2ca^5/b^5, and so on.
The waffle, now subdivided into infinitely many similar triangles. Along the side with length q, these hypotenuses have length c (from the original right triangle), c*2a^2/b^2, c*2a^4/b^4, c*2a^6/b^6, and so on.

Summing the segments along the line with length p gave you 2ac/b·[1 + (a/b)2 + (a/b)4 + (a/b)6 + …], an infinite geometric series that added up to p = 2abc/(b2a2). Meanwhile, summing the segments along the line with length q gave you c + 2a2c/b2·[1 + (a/b)2 + (a/b)4 + (a/b)6 + …], which added up to q = c(a2+b2)/(b2a2).

Other solvers, like Amy Leblang and Rohan Lewis, instead split the waffle into two right triangles, one of which was similar to the original triangle:

The same waffle shape shown above, but with additional labels. Now the top angles in the original right triangles (between sides c and b) have angle measure alpha, while the other acute angles have measure beta. The

In Rohan’s diagram, the unshaded triangle was similar to the overall construction. In particular, Rohan used the side lengths shown below:

The same waffle shape shown above, but with fewer labels, highlighting the similar triangles. The larger triangle has side lengths c, p and q. The smaller triangle has corresponding side lengths 2a^2/c, p-2ab/c and q-c.

Using the fact that these triangles were similar, that meant c/(2a2/c) = p/(p−2ab/c) = q/(qc). The equation between the first and second expressions allowed you to solve for p: p = 2abc/(c2−2a2). The equation between the first and third expressions allowed you to solve for q: q = c3/(c2−2a2).

For extra credit, you had to use these expressions for p and q to complete a proof of the Pythagorean theorem. Jackson and Johnson did this using the law of sines. Another way to prove the theorem was to equate the two above expressions for p. They had the same numerator, and equating their denominators gave b2a2 = c2−2a2, which could be arranged to the more familiar a2+b2 = c2

There are so many ways to prove the Pythagorean theorem. Thanks to Jackson, Johnson and this week’s solvers, we now have a few more!

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 Riddler,” and it’s in stores now!

Want to submit a riddle?

Email Zach Wissner-Gross at


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.


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