Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from many top-notch puzzle folks around the world — including you!

Recently, we started something new: Riddler Express problems. These are bite-size puzzles that don’t take as much fancy math or computational power to solve. For those of you in the slow-puzzle movement, worry not — we still feature our classic, more challenging Riddler.

You can mull both over on your commute, dissect them on your lunch break and argue about them with your friends and lovers. When you’re ready, submit your answer(s) using the links below. I’ll reveal the solutions next week, and a correct submission (chosen at random) will earn a shoutout in this column.

Before we get to the new puzzles, let’s reveal the winners of last week’s. Congratulations to 👏 Daniel Borup 👏 of Champaign, Illinois, and 👏 Phoebe Cai 👏 of Cambridge, Massachusetts, our respective Express and Classic winners. You can find solutions to the previous Riddlers at the bottom of this post.

First up, this week’s Riddler Express, a coin puzzle from Keith Wynroe.

You place 100 coins heads up in a row and number them by position, with the coin all the way on the left No. 1 and the one on the rightmost edge No. 100. Next, for every number N, from 1 to 100, you flip over every coin whose position is a multiple of N. For example, first you’ll flip over all the coins, because every number is a multiple of 1. Then you’ll flip over all the even-numbered coins, because they’re multiples of 2. Then you’ll flip coins No. 3, 6, 9, 12 … And so on.

What do the coins look like when you’re done? Specifically, which coins are heads down?

Submit your answer

If you need a hint, you can try asking me nicely. Want to submit a new Riddler Express? Email me.

And now, for Riddler Classic, more coins in this deadly board game puzzle from James Kushner of Sarasota, Florida:

While traveling in the Kingdom of Arbitraria, you are accused of a heinous crime. Arbitraria decides who’s guilty or innocent not through a court system, but a board game. It’s played on a simple board: a track with sequential spaces numbered from 0 to 1,000. The zero space is marked “start,” and your token is placed on it. You are handed a fair six-sided die and three coins. You are allowed to place the coins on three different (nonzero) spaces. Once placed, the coins may not be moved.

After placing the three coins, you roll the die and move your token forward the appropriate number of spaces. If, after moving the token, it lands on a space with a coin on it, you are freed. If not, you roll again and continue moving forward. If your token passes all three coins without landing on one, you are executed. On which three spaces should you place the coins to maximize your chances of survival?

Extra credit: Suppose there’s an additional rule that you cannot place the coins on adjacent spaces. What is the ideal placement now? What about the worst squares — where should you place your coins if you’re making a play for martyrdom?

Submit your answer

If you need a hint, you can try asking me nicely. Want to submit a new Riddler? Email me.

Here’s the solution to last week’s Riddler Express, which asked how many paths a lowly pawn could take across a chessboard to become a mighty queen. If the pawn begins on the e2 square and heads for the e8 square, there are 141 paths it could take.

The number of paths leading to various squares, and culminating with 141 possibilities at e8, is illustrated below, from the puzzle’s submitter Nik King:

To calculate these, first recognize that the pawn can move in one of three directions with its first move. Then, for each new square farther up the board, the possibilities are the sum of the possibilities on the previous squares from which it can reach that new square. For example, the number of possible paths to reach e5 is the sum of the paths that reach d4, e4 and f4 (2+3+2=7). Or the number to reach d6 is the sum of those that reach c5, d5 and e5 (3+6+7=16). Keep going in that fashion, and eventually you’ll reach 141 on e8.

(Note: If you counted the pawn moving forward two squares with its initial move as distinct from its moving two individual squares, then there are 160 paths. Feeling generous, I gave full credit for either approach.)

A special shoutout to Riddler Nation’s youngest (so far) citizen, Harry. The 7-year-old’s father sent along his solution, shown in part below, which he calculated using techniques that he learned watching math videos over the summer. Bravo, Harry!

And here’s the solution to last week’s Riddler Classic, concerning the paths taken around a board by various chess pieces, standard and unorthodox. The longest path a knight can take without that path intersecting itself is 35 moves. For a camel (which moves like a knight, except 3 squares by 1 square), it’s 17 moves; for a zebra (3 by 2), it’s also 17; and for a giraffe (4 by 1), it’s 15.

The longest knight paths are known for boards up to size 9-by-9, and those solutions can be found in The On-Line Encyclopedia of Integer Sequences.

This is a tough problem — there are lots and lots of paths a chess piece can take! — and one that’s best approached with computation. Famed computer scientist Donald Knuth tackled the knight problem in his book “Selected Papers on Fun and Games,” and there’s a lengthy article on “knight graphs” in Wolfram MathWorld. All the correct solutions I received took a programmatic approach of one kind or another. For one example, Laurent Lessard was good enough to share his Python code and describe his approach.

Another correct solver, Robbie Ostrow, described his solution this way: “Recursive backtracking! Only took about 30 hours …” Thirty hours very well spent, Robbie!

As ever, Laurent also provided a nice illustration of his solution paths:

Zach Wissner-Gross took a look at zebras moving around on 11-by-11 boards and figured out which starting squares are more amenable to long paths:

Recursively tracking nonintersecting "zebras" on 11×11 chessboards for this week's #Riddler. Greener starting points are better. @ollie pic.twitter.com/pUs8xdEYYY — Zach Wissner-Gross (@xaqwg) October 9, 2016

Zebras like the corners.

Elsewhere in the puzzling world:

Have a wonderful weekend!