POCO is 4595 and MUCHO is 68925.

There are 15 POCOs, so the sum of each column is a multiple of 15, and hence (without the carry from the column to its right) ends in a 5 or a 0. The rightmost column has no carry, so the O in MUCHO stands for 0 or 5. Which is it?

• If it’s 0, that means that the sum of all the Os in the rightmost column is 0, and that there’s no carry into the tens column. If there’s no carry, we need to find a digit for C that when multiplied by 15 produces a two-digit answer (the CH in MUCHO) that starts with that digit (since the hundreds column in the sum is otherwise zero). The only digit that will do that is 1 (1 × 15 = 15). But if CH is 15 then there’s no carry into the thousands column, which means that U must be 0 or 5 (15 × P). And we’ve already assigned 0 to O and 5 to H. So this can’t be right.
• If it’s 5, then the carry into the tens column is 7 (15 × 5 = 75). What is C? Every possibility from 0 to 8 produces conflicts or requires C to have two different values, but C = 9 produces a sum in the tens column of 142 (15 × 9 plus the 7 carry). In that case H = 2 and 14 is carried into the hundreds column. We know that O is 5, so that column now totals 89 (5 × 15 + 14). C is 9 and 8 is carried into the thousands column, where it’s added to 15 Ps. Here every possibility except P = 4 creates duplications or conflicts, so U is 8 and M is 6.

“Because O must equal 5 and C must equal 9 and P must equal 4, I believe the solution is unique,” writes NYU computer scientist Allan Gottlieb, who conducts the puzzle column. “I like these alphabetical number problems; but, cannot recall seeing this one before.”

1. Bc8! Kg4 2. Re4#

From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

Consider all the possible divisions, and in each case count up the total number of cases E in which a member has an enemy in his own house. The division in which E is minimized must have the required property. If it didn’t, that would mean that some member has at least two enemies in his own house. In that case he’d have one enemy at most in the other house, and could reduce E further by moving to the opposite house. That’s a contradiction, so the required task must be possible.

The probability is 100 percent. First suppose that Pile A is all red and Pile B all black. This fulfills the condition. Now move one red card from A to B. In order to keep 52 cards in each pile, we must also move a black card from B to A, so this new arrangement also fulfills the condition. And so on: So long as there are 52 cards of each color and 52 cards in each pile, the number of red cards in one pile must equal the number of black cards in the other. So you don’t have to view any cards at all.

See Alcohol Problem.

1. Nd5! Kxd5 2. Rb5#

From O.A. Brownson Jr., Chess Problems, 1876.

a + b = 1/4, b + c = 1/2, and a + d = 1/2. Triangles a and c are similar, and c has twice the linear dimensions of a, so c = 4a. That’s enough to work out the areas: a = 1/12, b = 1/6, c = 1/3, and d = 5/12.

From University of Toronto mathematician Ed Barbeau’s After Math (1995).

The surprising 1. Qxh8! leaves Black with no options. If he captures on c4 or d3 with his knight, then White recaptures with his king, giving mate. Any other response allows mate along the first rank or first file.

From Prager Tagblatt, Jan. 17, 1904.

1. Qd2!, waiting.

From Deutsche Schachzeitung, 1905.

Triangles CAM and ABN have the same area. Rotating triangle ABC 120° takes CAM into ABM1. Since ABN and ABM1 have the same area and both N and M1 fall on BC, N = M1. Since rotating CM produces AN, the angle between them is 120°, or 180° – 120° = 60°, which is the same.

(Posed by V. Proizvolov in Math Horizons, Spring 1994.)

1. Re3! wins at once: If the king takes the rook then 2. Qd2# is mate; if the pawn moves then 2. Re4#; and if the knight moves then 2. Qe5#.

From Alain C. White, Sam Loyd and His Chess Problems, 1913.