The Martian is a book by Andy Weir about how potatoes will become the first food crop on Mars. Alright, it’s about more than that, but anyone who has read Weir’s novel knows potatoes get plenty of mentions. Anyway, the point is: This week, in honor of Mark Watney’s space spuds, we’re featuring a puzzle about Martian tubers.

Before we get to the puzzle, though, a brief announcement: This was my last week with io9 (I’ve written about my departure here), which makes this the last installment of Sunday Puzzle—at least for the foreseeable future. Writing this column has been one of the best parts of my time here at io9, and it’s been a pleasure watching you all work these brain teasers out in the comments. You folks are some formidable problem solvers; it’s been an honor to serve these up and watch you plow through them week by week.

Sunday Puzzle #46: The Potato Paradox

This is an oldie but a goodie that, for reasons mentioned above, recently came to mind while reading Andy Weir’s The Martian. What I really love about it is its simple presentation; unlike a lot of the puzzles I’ve featured here, this one is easy to memorize, and can be straightforwardly posed in just about any situation. But for as simple as this puzzle reads, the solution almost always catches people off guard, and so it’s referred to as a paradox (a veridical paradox, to be exact). The problem reads as follows:

You have 100 pounds of Martian potatoes, which are 99 percent water by weight. You let them dehydrate until they’re 98 percent water. How much do they weigh now?

Solutions will work a little differently this go around: Next week, I’ll come back to highlight the correct responses in the comments section below, so check back at this URL in seven days’ time. (Or, if you’re reading this more than one week in the future, just scroll down to the comments!)

SOLUTION to Sunday Puzzle #45: A Devious Selection Task

Last week, I presented you with the four cards pictured below and tasked you with turning over as few cards as possible to verify whether the following statement is true: Every card with a vowel on one side has an even number on the other side:

The first person to provide the correct response was Y_Y_Zhed, who explained their reasoning as follows:

I’ll give it a go:

You have to flip over “A”, to make sure that it has an even number

You have to flip over “B” to make sure that it *doesn’t* have a vowel

You have to flip over “7” to make sure that it *doesn’t* have a vowel

My answer is 3 cards.

In other words: Every card but the one with “4” on its upward-facing side must be flipped. The “4” need not be flipped because turning the card over can neither prove nor refute the statement that “Every card with a vowel on one side has an even number on the other side.”

I cautioned you last week to be careful answering this puzzle if you’d seen a similar problem in the past. Here’s why: In most versions of this problem, it is explicitly stated that each card has a letter on one side and a number on the other. This puzzle, however, did not. If you, like many people did last week, answered that only “A” and “7” needed to flip, you likely assumed that a number had to appear on the obverse of “B.”

This puzzle and others like it are all variations on a logic puzzle developed by psychologist Peter Cathcart Wason in the 1960s, and are called, rather appropriately, “Wason selection tasks.” You’ll find variations of them all over the Internet, and a lot of them are just a tiny bit different from one another, so you can test yourself to make sure you’ve really got the reasoning behind them down pat. This is a good place to start.

And oh! I almost forgot! Last week, hoberiam gave an excellent lateral solution to this puzzle that I’d never seen before. Their answer was completely unexpected and, I think, totally valid given the phrasing of the puzzle:

I would flip over 0 cards. The puzzle doesn’t specify which sides. Instead of thinking about it as front side and back side, I instead look at the left and right sides of each card.

From there, I can state that B is the only card with a vowel on one side (left). And since it also has an even number on its other side (right), I can conclude the statement is true.