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.

## Riddler Express

From Lucas Jaeger comes a âflute-ifulâ challenge:

Youâre a contestant on the hit new game show, âYou Bet Your Fife.â On the show, a random real number (i.e., decimals are allowed) is chosen between 0 and 100. Your job is to guess a value that is *less than* this randomly chosen number. Your reward for winning is a novelty fife that is valued precisely at your guess. For example, if the number is 75 and you guess 5, youâd win a $5 fife, but if youâd guessed 60, youâd win a $60 fife. Meanwhile, a guess of 80 would win you nothing.

What number should you guess to maximize the average value of your fifing winnings?

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

In a world of sudokus, KenKens and kakuros, Barbara Yew offers a different sort of number puzzle:

There are eight three-digit numbers â each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens, and ones places, respectively, are shown in the bottom row.

Can you find all eight three-digit numbers and complete the table? Itâs a bit of a mystery, but Iâm sure you have it within you to hunt down the answer!

The solution to this Riddler Classic can be found in the following column.

## Solution to last weekâs Riddler Express

Congratulations to đ Jacob Kopczynski đ of San Francisco, California, winner of last weekâs Riddler Express.

Last week, you were slicing one big square into smaller squares (not necessarily of equal size), so that the smaller squares didnât overlap, while still making up the entire area of the big square.

What whole numbers of squares could you *not* have sliced the big square into?

Many solvers began by trying out small numbers of squares. Of course, it was possible to slice the big square into one square by simply leaving it alone. The next smallest number of squares you could get was 4, each a quarter of the big square. That meant getting either two squares or three squares was impossible. (A rigorous proof of this is left as an exercise for you, the reader â hah!)

Another possible number was 9, since that would be a 3-by-3 array of equally sized squares. Similarly, any square number was possible. But what about numbers *between* the perfect squares?

As noted by solver America Masaros, if it was possible to create N squares, it was also possible to create *N*+3 squares by taking any undivided square and slicing it into four equally sized squares. This effectively replaced one square with four new squares â a net gain of three squares. That meant any number that was 4 or 9 plus a multiple of 3 (i.e., of the form 4+3*k* or 9+3*k*, where *k* is a whole number) was also possible, ruling out a whole bunch of numbers.

But there were still a few numbers less than 9 that had to be checked: 5, 6 and 8 (7 was 3 more than 4, so it was possible). While 5 was not possible (again, you are welcome to prove this!), both 6 and 8 *were* possible using unequally sized squares, as shown below:

Indeed, this strategy with one larger square and an elbow of smaller squares around it worked for any even number greater than 2.

Finally, since 8 was possible, any number that was a multiple of 3 greater than 8 was also possible, again by picking one of the squares and slicing it into four equal squares. That meant the only numbers that were *not* possible were **2, 3 and 5**.

Meanwhile, Vince Vatter (of Battle for Riddler Nation fame) recognized this problem from one of his colleagueâs books:^{2}

It remains unclear why a value of 14 was chosen for this exercise, rather than the actual lower bound of 5.

At least one solver extended the puzzle further, looking at how a cube could be partitioned into smaller cubes. But letâs save that for a future riddle!

## Solution to last weekâs Riddler Classic

Congratulations to đ Eilon đ of Chicago, Illinois, winner of last weekâs Riddler Classic.

Last week, Robin of Foxley entered the FiveThirtyEight archery tournament. She was guaranteed to hit the circular target, which had no subdivisions â it was just one big circle. However, her arrows were equally likely to hit each location within the target.

Her true love, Marian, had issued a challenge. Robin had to fire as many arrows as she could, such that each arrow was closer to the center of the target than the previous arrow. For example, if Robin fired three arrows, each closer to the center than the previous, but the fourth arrow was farther than the third, then she was done with the challenge and her score was *4*.

On average, what score could Robin have expected to achieve in this archery challenge?

First off, a surprising and subtle fact about this riddle was that the geometry (in this case, a circle) didnât matter. What did matter was how close each arrow was to the center *relative to the other arrows*. That meant the target could have been a square, a line segment or even a sphere â the answer would be the same.

A good first step was then to restate the problem and forget about the geometry: If you pick random values between 0 and 1 uniformly â each representing the relative distance of an arrow to the center â how many consecutive decreasing values would you expect (plus one, for the arrow that broke the streak)?

Solver Balthazar Potet approached this by thinking about the values for the first *N* arrows Robin fired and the probability theyâd result in a score of *N*. With any *N* values, there were *N*! ways to order them. For Robin to have a score of *N*, the smallest value couldnât have been in the *N*th position, since it had to be greater than the previous value. And when each of the other *N*â1 values occurred in the *N*th position, there was exactly one way to order the remaining values so that they formed a decreasing sequence. So of the *N*! orderings, *N*â1 resulted in a score of *N*, meaning the probability was (*N*â1)/*N*!

From there, you had to use these probabilities to compute an average score, which you could find by multiplying each score by its probability and then adding up all those products. The probability Robin scored 2 was (2â1)/2!, or 1/2, which meant a score of 2 contributed 2Â·1/2, or 1, to her average score. The probability Robin scored 3 was (3â1)/3!, or 1/3, which meant a score of 3 contributed 3Â·1/3, or 1 (again!), to her average score. In general, the probability Robin scored *N* was (*N*â1)/*N*!, which meant a score of *N* contributed *N*Â·(*N*â1)/*N*!, or 1/(*N*â2)!, to her average score. Since *N* was at least 2 â meaning Robin fired at least two arrows â her average score was 1/0! + 1/1! + 1/2! + 1/3! + âŠ, a sum that converges to ** e**, which is

**approximately 2.71828**. Huzzah, another riddle whose answer was a famous mathematical constant!

For extra credit, you had to calculate Robinâs average score when the target had 10 concentric circles, whose radii were 1, 2, 3, etc., all the way up to 10 (the radius of the entire target). This time, Robin had to fire as many arrows as she could, such that each arrow fell within a smaller concentric circle than the previous arrow.

Here, the geometry of the target was relevant, since there was now a nonzero probability that consecutive arrows could fall within the same ring. The chances that any given arrow landed in one of the rings (from the smallest ring to the largest) were 1 percent, 3 percent, 5 percent, 7 percent, 9 percent, 11 percent, 13 percent, 15 percent, 17 percent and 19 percent.

Having the discrete rings (perhaps counterintuitively) made the problem more complex, but several solvers persisted. Emma Knight was able to set up and solve a system of 10 equations, finding that the average number of arrows was **approximately 2.5585**. Josh Silverman was further able to come up with a closed formula for the solution and a precise rational result.

Meanwhile, solvers like Paulina Leperi and Angelos Tzelepis (whose results are shown below) approximated the answer by simulating many arrows.

That was the answer when there were 10 rings. As the number of rings increased, the average score also increased, approaching *e* in the limit of infinitely many rings.

By now, I bet youâre curious how Robin of Foxley actually performed at Marianâs archery challenge. Robin was such a good archer that she was able to fire off two arrows that were *exactly* the same distance from the targetâs center. Marianâs mind was blown by this probability-zero event, and they lived happily ever after.

## Want more riddles?

Well, arenât you lucky? Thereâs a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. Itâs called âThe Riddler,â and itâs in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com