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.

Eleven. For any integer n greater than that, the possible remainders when n is divided by 3 are 0, 1, and 2. If the remainder is 0, then we can get to n by scoring field goals. If the remainder is 1, then n – 13 has remainder 0 and we can reach n by scoring one touchdown and some number of field goals. And if the remainder is 2, then n – 14 has remainder 0 and we can get to n by scoring two touchdowns and some number of field goals. Put simply, all integers n greater than 11 can be written in the form n = 7a + 3b for some nonnegative integers a and b.

Altogether there are 2n + 2 people at the party. None shakes hands with themselves or their partner, so the maximum number of handshakes that one person could have participated in is 2n, and the minimum is 0. Anna receives a different answer from each of the 2n + 1 people she asks, so the answers she gets must be 0, 1, 2, …, 2n.

Let the person who shook hands 2n times be called Max and the person who shook hands 0 times be called Minie (making no assumptions about gender). Max must have shaken hands with everyone but his or her own partner. Since Max has shaken hands with Anna, Max cannot be Bert and Max’s partner must be Minie.

Now remove Max and Minie and all their handshakes. This decreases by 1 the number of handshakes that each of the others participated in, and leaves us facing essentially the same problem but with n – 1 other couples. If we repeat this process, eventually we’re left with Anna and Bert as the only two people remaining. In that situation, neither of them shakes hands. And since one handshake disappeared as each of the other n couples was removed from the problem, both Anna and Bert have shaken hands n times.

(Paul Belcher, “But What Colour Are Anna’s Eyes?”, Mathematical Gazette 87:509 [July 2003], 322-323.)