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David Dewan came up with this solution (also found by Richard Lipes on Wikipedia): Let N be any integer not divisible by 2 or 5. Consider repunits R1 = 1, R2 = 11, R3 = 111, …, RN+1 and their remainders by modulus N. There are the most of them N different residues, so a set of N + 1 residues modulo N must contain at least one repeat. assume RAND mod N and RB mod N with B > AND are the same. Then (RB – RAND) mod N = (RB.A10AND) mod N = 0. Since N and 10 are relatively prime numbers, N shares RB.A.
S/O4. Frank offers this sudoku problem:
Many readers have successfully tackled it. For anyone stuck, here’s the answer key:
“Ten years ago I was doing what is now called AI,” writes Richard Marks ’58. He noted that trying all possible repetitions of the sudoku problem “will tie up your Cray for a week. I personally wrote the rules and coded this little AI program that solves every Sudoku by the time you release the “Solve” 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 checkmate in twos (see below).
Frank Model ’63 says that there are two cases to consider: when Black can rook and when Black cannot. If Black can rook, then Black’s last move must have been Nc7-c5. In this case white can take passing byPb5-c6. If Black gives, then White plays Pb7#. If Black can cast but makes some other move, then White plays Rf8#.
If Black cannot cast, then White plays Ke6. Black cannot escape checkmate on the next move when White plays Rf8#.
Steve Gordon noted that this problem is illustrated by three particular chess moves. The following is adapted from his analysis with his algebraic chess notation.
Black last moved either the king, rook or c5. If c5, then 1. bxc6 (passing by aka epic). If Black can still surrender the queenside (1. Kc8 & Rd8, also known as OOO), then 2. b7#. If black can’t, his king is still trapped at rank 8, so after any move by black, 2. Rf8#. If black moved the king or rook last, passing by not possible for White, but Black cannot rook either, so 1. Ke6 also traps Black’s king at rank 8, and after any move by Black, 2. Rf8#. Note: For the construction of this puzzle, the white knight on g7 is the result of an insufficient promotion on h8.
