Logic Masters Deutschland e.V.

Eliminate the impossible

(Eingestellt am 16. September 2025, 23:21 Uhr von PjoeterBliep)

Many thanks to sanabas for testing several broken versions of this puzzle. This one is not broken, I promise!

When you have eliminated the impossible,

whatever the remainders,

no matter the modulus,

must be the solution

Rules:

Normal sudoku rules apply: Place the digits 1-9 once each in every row, column and 3×3 box.

Bouncing Modular X-Products: The notation A (mB) at an arrow indicates that A is the product of the first N digits in the direction of the arrow, up to a multiple of the modulus B, where N is the first digit the arrow points at. If N is larger than the diagonal an arrow points at, the reflected diagonal is used as well.

Example: The clue in row 7 says, if r8c9 is a 4, then the product of the 4 digits at r8c9, r9c8, r8c7 and r7c6 must be equal to 8 + 15*k for some integer k.

Sudokupad

Lösungscode: Row 2 (left to right, no spaces)

Zuletzt geändert am 17. September 2025, 09:10 Uhr

Gelöst von sanabas, StefanSch, zeniko, SKORP17
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Kommentare

am 17. September 2025, 15:57 Uhr von zeniko
Interesting ruleset and about as tricky as it looks (definitively harder than 3* for me, and that was using a calculator). Thanks for sharing.

am 17. September 2025, 14:45 Uhr von StefanSch
Verry nice, but more math than a puzzle ;-)

am 17. September 2025, 09:10 Uhr von PjoeterBliep
Add nice quote

am 16. September 2025, 23:51 Uhr von sanabas
Novel rule idea leads to some fun logic

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ID:000P7V

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Lösungscode:

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