The test makers were expecting the answer (C), but several students pointed out that in fact (A) is correct — the shapes fit together to produce a surprisingly simple object with only five faces.

This is easiest to see if you imagine two square pyramids side by side: The tetrahedron fits precisely between them to make a smooth-sided “pup tent”:

If you have polyhedral dice, try fitting the face of a 4-sided to an 8-sided die; you’ll find that the adjoining sides are coplanar.

It’s the ace of diamonds.

If Val couldn’t name the card knowing only its value, then it must be an ace, a queen, a 5, or a 4, as these are the only values that appear more than once.

If Colin knew that Val couldn’t name it, it must be red, as all the red cards have values (A, Q, 5, 4) that appear more than once.

When Val understands that the card must be red, she can exclude from her list of possibilities the queen, the 5, and the 4, as each of those cards appears in a black suit.

That leaves only the ace of diamonds.

Adapted from Julian Havil’s Impossible?, 2008.

UPDATE: Wait, this is cooked. Havil gives the problem but not the solution, and my solution above isn’t valid. Thanks, Ben and Andrew, for pointing it out. If anyone can find a valid solution, let me know and I’ll post it here. I don’t think one is possible.

FWIW, this is based on a more famous “impossible puzzle,” which is soluble but spectacularly hard.

UPDATE 2: In Canada or England, does the word “colour” mean “suit”? If so, the problem is soluble. It seems to have appeared originally on page 14 of a paper [PDF] by University of Victoria philosopher Audrey Yap. The first two steps above are right — Val can deduce that the card is red and must be A, Q, 5, or 4. The fact that she’s able to say “I know the card now” at this point means it can’t be an ace (because that value appears in both red suits). And the fact that Colin can then say “I know it too” means that it must be the five of diamonds. So, from the top:

Val says, “I don’t know what the card is.” (She knows it’s a 5, but there are two 5s.)

Colin says, “I knew that you didn’t know.” (He knows it’s a diamond, and thus an A or 5; but as each of these values appears more than once in the deck, he knows that Val won’t have enough information yet to identify the suit.)

Val says, “I know the card now.” (Colin’s remark has just told her that the card is red, because all the red cards in this deck have values that appear in multiple suits. And the only red 5 is 5♦.)

Colin says, “I know it too.” (He knows that Val was looking for an A or a 5, and that learning the card’s color was helpful. Thus she can’t have been looking for an A, as both aces are red. That leaves 5♦.)

So, if “colour” means “suit,” the answer is 5♦. If “colour” means “color,” then I still think it’s insoluble. Thanks to the many smart people who wrote in about this, especially Helge.

White: To mate, 1. Rb3#. To self-mate, 1. bxc8, promoting to a black king, pinning the pieces on b8 and c7 and leaving Black with only 1. … Kxc3#.

Black: To mate, 1. … Kxc3#. To self-mate, 1. … Nxd5+, forcing 2. Nxd5#.

“Notwithstanding the memo, many young solvers (and old ones too) have had a hard wrestle with this production, but all to no purpose, it was the first of its kind published. The consequence was that many communications were received by the composer doubting his sanity, denouncing the problem as a fraud and threatening that if they caught him on the other side of the channel they would give him a taste of their wrath. The receiver of these communications thought that the channel was a mighty fine affair!”

From Thomas B. Rowland, Chess Fruits, 1884.

Our life is but a pilgrimage,
From birth unto the bier,
We journey onward as the Knight,
Our pathway never clear.
O’er chequered course of sixty-four,
Through ways that lead astray,
Now bright, then dark, we travel on,
Till darkness veils our day.

A man’s birth year plus his age always gives the current year; so does the sum of his hiring year and his time on the job. So here the “magic” total, 3888, is just twice 1944.

This works for any group of people — as long as they’re all still alive and working, these values will add to a miraculous constant. Here’s the current Supreme Court:

Their magic total, sure enough, is just twice 2009.

“If W and E were stationary points, and W, as at present, on your left when advancing towards the N, then, after passing the Pole and turning round, W would be on your right, as stated. But W and E are not fixed points, but directions round the globe; so wherever you stand facing N, you will have the W direction on your left and the E direction on the right.”

What piece did Black move last? It wasn’t the pawn on h4, which must have walked over from e7. The only other possibility is the pawn on c5, which must have moved from c7.

Then where did the white pawn on c6 come from? There’s only one possibility: White is midway through an en passant capture. That’s the only way this position could have arisen even momentarily in a legal chess game.

So now White just completes the capture, removing the black pawn on c5, and that’s mate:

From Thomas B. Rowland, Chess Fruits, 1884.

The conjurer simply adds up the digits in the final figure. If the result has multiple digits, he adds those, and so on until he reaches a single digit:

4 + 5 + 6 + 0 + 2 + 0 + 7 + 0 + 8 + 6 + 5 + 6 + 9 = 58
5 + 8 = 13
1 + 3 = 4

That digit is the number from step 1.

It works because all multiples of 9 eventually produce 9 when their digits are added as above; the procedure asks you to create two multiples of 9 (in steps 2 and 4) and to combine their digits and add your own. Adding the digits will thus yield 9 plus the number chosen (here, 13), and of course adding those digits will yield the number itself.

The business with the father’s first name, etc., is just smoke — any numbers at all can be used in those positions. The arithmetic will always produce the right answer.