Puzzling Asked on January 12, 2021
This is a Kyudoku Puzzle. I have made the rules myself.
There will be numbers in a grid. You just have to circle nine unique numbers (1 to 9) such that each row and column has sum of 9 or less. In some puzzles, one or more circled numbers may already be given.
Here is the real puzzle.
9 | 1 | 2 | 7 | 6 | 7 |
9 | 9 | 5 | 5 | 4 | 1 |
5 | 7 | 9 | 4 | 3 | 8 |
4 | 6 | 8 | 5 | 1 | 9 |
9 | 5 | 8 | 7 | 2 | 8 |
1 | 8 | 6 | ② | 2 | 8 |
Note:- This is not a puzzle of my own. There are puzzles similar to this you will find in the internet, but it has not been introduced that well in PSE before.
The 2 is already circled, so we can grey out all the other 2s in the grid.
There are only three ungreyed 8s in the grid now (on the fourth and fifth rows), so one of them must be circled.
There are only three ungreyed 9s in the grid now (top left corner), so one of them must be circled.
There are only three ungreyed 6s and three ungreyed 7s in the grid now.
Now we know that
Correct answer by Rand al'Thor on January 12, 2021
Here is the answer you are looking for:
First of all some obvious deductions, the crosses are the excluded numbers:
Then I tried:
Then with this hypothetical:
Then I figured out
Lastly:
Answered by Smartest1here on January 12, 2021
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