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Sisyphus Prime

(Published on 10. July 2024, 08:31 by PhysicistFromFunen)


2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
All one- and two-digit primes are conveniently listed above. Here are the rules:
  1. Normal Sudoku rules apply.
  2. Sisyphus is cursed to roll a boulder uphill on the crimson lines. Upon reaching the end of a line the boulder falls vertically onto the base of the next - alas, Sisyphus must climb another line. Box borders divide the crimson lines into segments. Digits on each segment must make up a prime number when read along Sisyphus's path. The sum of all such primes on a line must itself be prime.
  3. The grey lines are palindromes, i.e. the Xth cell from one end must have the same digit as the Xth cell from the other end.
  4. The sum of the U-shaped palindrome must be double the sum of the digits appearing on the palindromes making the P, i.e. R3C5 and R4C6 count only once towards the sum of the P.
The puzzle is also available online in the CtC App. I will greatly appreciate any feedback! Have fun solving!

Solution code: Digits in rows 1 and 8.

Last changed on on 10. July 2024, 10:27

Solved by Megalobrainiac, StefanSch, SKORP17
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Comments

on 10. July 2024, 10:27 by PhysicistFromFunen
Clarified rule 4. Thank you StefanSch for pointing out the misleading phrasing.

Last changed on 10. July 2024, 09:50

on 10. July 2024, 09:38 by StefanSch
I thougt the 4. rule means: R7C1+R8C2+R9C3+R8C4+R7C5+R6C4+R5C3 = 2*(R3C5+R4C6+R5C7 + R3C5+R2C6+R3C7+R4C6)
But it is: R7C1+R8C2+R9C3+R8C4+R7C5+R6C4+R5C3 = 2*(R3C5+R4C6+R5C7+R2C6+R3C7) !

Difficulty:3
Rating:N/A
Solved:3 times
Observed:2 times
ID:000IV7

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