As usual in chess, let the bottom left-hand square be black. If we number the rows from the bottom and the columns from the left, then every black square is either in an odd-numbered row or an even-numbered column. That is, all the black squares can be found in rows 1, 3, 5, 7 and columns 2, 4, 6, 8.

Let the black squares in rows 1, 3, 5, 7 be marked A and the black squares in columns 2, 4, 6, 8 be marked B, and let the white squares in rows 1, 3, 5, 7 be marked C. Leave the rest of the white squares unmarked. Finally, let a, b, c denote the total number of pieces that sit on squares marked A, B, C, respectively. We want to show that a + b is an even number.

Each row contains an odd number of pieces. That means that a + c, the total number of pieces in rows 1, 3, 5, 7, is the sum of four odd numbers, which means it’s even. Likewise, b + c, the total number of pieces in columns 2, 4, 6, 8, is also the sum of four odd numbers and thus even. So

(a + c) + (b + c) = a + b + 2c

is even, and this implies that a + b is even.

(From V. Proizvolov, “Problems Teaching Us How to Think,” Quantum, January-February 2002, via Ross Honsberger’s Mathematical Delights, 2004.)

1. Qb7! and the queen or knight will mate.

From Deutsches Wochenschach, 1893.

TTT = T × 111 = T × 3 × 37, so either YE or ME is 37. Either way, E is 7.

T is a digit, and T × 3 is a two-digit number ending in 7 (it’s either YE or ME, whichever of these is not 37).

So T must be 9, which means that TTT = 999 = 27 × 37, and E + M + T + Y = 2 + 3 + 7 + 9 = 21.

UPDATE: Strictly speaking we can’t assume that one of the factors must be 37 — it could be 74, a two-digit multiple of 37. But then when we consider the other factor, the only candidate that yields a product with the pattern TTT is 12, which doesn’t fit. Still, it’s necessary to check this line to be sure the solution is unique. (Thanks, Jeff and Steven.)

Of the nine nonzero digits, three require three characters (1, 2, 6), three require four characters (4, 5, 9), and three require five characters (3, 7, 8).

This means that the last digit of the number we’re looking for must be zero, because any other digit would have counterparts of equal length. EIGHTY-ONE, for instance, is the same length as EIGHTY-TWO and EIGHTY-SIX.

(It’s true that there are some unusually named numbers between 10 and 20, but it turns out that only one of these has a unique number of characters — 17, with nine — and 42 also uses nine characters.)

The tens digit must also be zero, for similar reasons: 20, 30, 80, and 90 all use six characters; and 40, 50, and 60 use five. 10 uses three characters, but this matches 1, 2, and 6. 70 uses 7 characters, but so does 15.

The discussion above regarding the units digit applies also to the hundreds digit, so that too is zero, and the candidates we’re left with are 1000, 2000, 3000, 4000, and 5000. Of these, only 3000 has a unique length, with 14 characters.

(Proposed in Pi Mu Epsilon Journal Spring 1985 by Harry Herein; solved in PMEJ Spring 1986 by Bob LaBarre.)

Since y is y% of z, z must be 100. The situation is possible if and only if y is 100 as well.

The key is the 4 of hearts. There is no 3 or 5 present, so the 4 of hearts can’t be part of a straight. There are only three other hearts, so it can’t be part of a flush. It’s the only 4 in the set, so it can’t be part of a full house. And the set contains no four cards of the same value, so the four of hearts can’t be the fifth card accompanying four of a kind.

1. Qh3! Ke4 2. Rc4#

From the Dubuque Chess Journal, 1873.