Logic Masters Deutschland e.V.

Powers of Two

(Published on 9. September 2025, 08:00 by Kaktuslav)

Link to SudokuPad: https://sudokupad.app/pvqdcm5cnt



Rules: Normal sudoku rules apply. Logarithmic arrows: Bulbs contain pairwise different (i.e. non-repeating) powers of two (1-, 2-, or 3-digit numbers written in decimal). If a bulb contains 2^x (2 to the power x), the digit on the attached arrow must be x.

Solution code: Row 4

Last changed on on 9. September 2025, 18:17

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Comments

yesterday, 03:55 by dzamie
Math :)

on 9. September 2025, 18:35 by Franjo
From a setter’s perspective you did very well in finding a way to put those nine L.A.‘s into the grid that leads to a unique solution. From a solver’s view it was extremely easy to fill all the bulbs and arrows and get an easy-to-solve standard sudoku. Anyway, thank you for sharing this interesting - experiment(?).

on 9. September 2025, 18:30 by Supertaster
Each bulb, read as a single number left-to-right, is a power of two. The power is pointed to by the arrow.

For example, 2^7 = 128, so you might have a bulb that reads 128 and points to a 7.

on 9. September 2025, 18:26 by sfield
"pairwise different" would make sense to distinguish between {2,4} and {4,2}. But one of those is a valid {x,2^x} pair and the other isn't. So including the term pairwise just makes it confusing. I tried solving it for a while before realizing each power had to be different and then it became easy.

on 9. September 2025, 18:17 by Kaktuslav
Clarified rules

on 9. September 2025, 18:09 by Dermerlin
Is it really each bulb's SUM that is a power of 2???
I'd guess it's the number writeen in the bulb and not the sum.

on 9. September 2025, 15:42 by Grumpy
Yeah, the rules are a bit hard to grasp at first.

But essentially:
- Each bulb has a different sum
- Each bulb's sum is a power of 2
- The number on the arrow is the exponent to raise 2 to, to get the sum in the bulb. For example: 4<-(16), since 2^4 = 16

on 9. September 2025, 13:57 by Lego7656
pairwise? what pairs?

Difficulty:1
Rating:88 %
Solved:61 times
Observed:0 times
ID:000P15

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Solution code:

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