1. a5! leaves the black king with no moves. Black must move the pawn, whereupon White captures it, giving mate. (If Black moves the pawn two squares, White can capture it en passant, giving the same result.)

From Rivista Scacchistica Italiana, 1907.

The Xs in the left column produce a 1-digit sum, so X = 1, 2, 3, or 4. But looking at the right column, X is the sum of Y and Z, which are distinct numbers, so the smallest possible value of X is 1 + 2 = 3. So X must be either 3 or 4. That means (looking at the left column again) that Y must be at least 6. But Y and Z are summed in both the center and the right columns, producing different digits. So Y and Z must total at least 11, and that means that a 1 is being carried into the total X + X in the left column, producing a total for that column (and a value for Y) of either 3 + 3 + 1 = 7 or 4 + 4 + 1 = 9.

The fact that we’re carrying a 1 into the center column shows us that X + 1 = Z. We’ve established that X is 3 or 4, so Z must be 4 or 5. So:

X = 3 or 4
Y = 7 or 9
Z = 4 or 5

A bit of trial and error shows that X = 4, Y = 9, Z = 5.

08/18/2017 UPDATE: There’s a simple solution that I overlooked:

The tens place shows Z + Y = Z. So either Y is 0 or Y is 9 and we’ve carried a 1 from the ones place. And Y can’t be 0 because it’s the first digit of the solution. So Y is 9.

The hundreds place shows X + X = 9. That’s possible only if X is 4 and we’ve carried a 1 from the tens column.

And the ones place shows Y + Z = X; that is, 9 + Z = 4. So Z is 5.

Also, my argument that X can’t be 1 or 2 isn’t valid — the X in the first column could be the ones digit of some two-digit sum.

Thanks to everyone who wrote in about this; sorry I couldn’t reply to each of you individually.

06/01/2018 Yet another solution:

• From the first column we know that X < Y, since there is no 1 carried forward.
• This means the third column is effectively Y + Z = X + 10 (a 1 has to be carried forward, per the observation above).
• Now the second column gives this equation: Z + Y + 1 = Z + 10 (again a 1 must be carried forward, since Y can’t be zero).
• Hence the first column gives the equation: X + X + 1 = Y.

Solving the three equations gives the three variables. (Thanks, Tom.)

One. Take one coin from the first stack, two from the second, and so on, and weigh this collection together. The number of grams by which the total is light is the number of the counterfeit stack.

41312432 (or its reverse, 23421314).

1. Rd1! and the queen will mate.

1. Rf2! Bxf2+ 2. Kxf2#

Suppose the man normally reaches the crossing, B, at 8 a.m. Then on a typical day he’d reach A, 6 miles before the crossing, at 7:30 a.m.

When the man is 25 minutes late, he’ll reach B at 8:25 a.m. and A at 7:55 a.m.

Hence the train takes 5 minutes to travel the 6 miles from A to B.

Therefore its speed is 72 mph.

Cheryl’s birthday is July 16.

Albert’s first statement is unsurprising — he knows only the month of Cheryl’s birthday, and each month on the list includes several possible days. But the fact that he knows that Bernard doesn’t know it is important — it tells us that the month can’t be May or June, because both of those months contain unambiguous dates. For example, if Cheryl’s birthday were May 19, then Bernard would know immediately that she was born in May, as that’s the only month whose 19th day appears on the list. Similarly, if Bernard knew the day of Cheryl’s birthday was the 18th, he’d know immediately that she was born in June. The fact that Albert can exclude these possibilities shows that he knows she wasn’t born in May or June; she must have been born in July or August.

Bernard’s statement shows that he’s followed this and now knows that the month must be either July or August. And the fact that he now knows the full birthday shows that Cheryl was born on the 15th, 16th, or 17th — if she’d been born on the 14th then he’d still be uncertain, as both July 14 and August 14 fall on the list.

Albert has narrowed the list of possibilities to July 16, August 15, and August 17. As he now declares that he knows the birthday unambiguously, it must be July 16 — if it fell in August he’d have no way to distinguish between August 15 and August 17.

No, there isn’t. Color the cells as shown:

Any path that the king takes must alternate between dark and light rooms, so the king can succeed only if there are an equal number of each (or if their numbers differ by 1). In the diagram there are 12 dark rooms and 10 light rooms, so there’s no way to visit all of them on one connected tour, unless the path travels outside the palace.

(From Tanton’s “A Dozen Questions About a Triangle,” Math Horizons 9:4 [April 2002], 23-28.)