Puzzle Corner September/October 2024 bonus solutions

Here are solutions for the three bonus problems that appeared in the September/October 2024 Puzzle Corner column we guest edited. Solutions for S/O2, S/O4, and S/O6 are below; those for S/O1, S/O3, and S/O5 can be found here.

S/O2. Frank notes that a repunit R_k is a decimal integer consisting of the digit 1 repeated k times, with k > 0. For example, R₁ = 1, R₂ = 11, R₃ = 111, etc. Let N be any integer not divisible by 2 or 5. Prove that there is a repunit divisible by N.

David Dewan came up with this solution (which Richard Lipes also found on Wikipedia): Let N be any integer not divisible by 2 or 5. Consider repunits R₁ = 1, R₂ = 11, R₃ = 111, …, R_N+1 and their residues modulo N. There are at most N different residues, so the set of N + 1 residues modulo N must contain at least one repeat. Assume R_A mod N and R_B mod N with B > A are the same. Then (R_B – R_A) mod N = (R_B-A10^A) mod N = 0. Since N and 10 are relatively prime, N divides R_B-A.

S/O4. Frank offers this sudoku problem:

Many readers tackled it successfully. For anyone who got stuck, here’s the answer key:

“Ten years ago I was doing what is now called AI,” Richard Marks ’58 writes. He noted that trying all possible iterations of a sudoku problem “will tie up your Cray for a week. So I personally wrote the rules and coded this little AI program that solves any sudoko by the time you have released the “Solve It” button. Since I wrote the program from scratch, I guess you can say I solved S/O4.”

S/O6. On behalf of the MIT Chess Club, Justin Zhou ’25 asked how White can play and mate in two (see below).

Frank Model ’63 says there are two cases to consider: when Black can castle, and when Black cannot. If Black can castle, then Black’s last move had to be Pc7-c5. In this case White can take en passant, Pb5-c6. If Black castles, then White plays Pb7#. If Black can castle, but makes some other move, then White plays Rf8#. 

If Black cannot castle, then White plays Ke6. Black cannot escape mate on the next move when White plays Rf8#.

Steve Gordon noted that this problem illustrates the three special chess moves. The following was adapted from his analysis with his algebraic chess notation. 

Black last moved either a king, rook, or c5. If c5, then 1. bxc6 (en passant a.k.a e. p.). If Black can still castle queenside (1. Kc8 & Rd8, a.k.a O-O-O), then 2. b7#. If Black cannot, its king is still trapped on rank 8, so after any black move, 2. Rf8#. If Black last moved a king or rook, en passant is not possible for White, but Black can’t castle either, so 1. Ke6 also traps the black king on rank 8, and after any Black move, 2. Rf8#. Note: To create this puzzle, the white bishop on g7 resulted from an underpromotion on h8.