Brainlet here. How to solve this fucker?

Brainlet here. How to solve this fucker?

Attached: 3.jpg (1080x1077, 285K)

add up all the grey squares and divide by total blue numbers

6th column middle box:

4,6,7

why

You put numbers in the large boxes. You cant repeat the same number in any line or row.

You have a trio of 1,3 and 8 in 3 boxes so you can eliminate any other numbers from those boxes. That leaves you with the top right of the middle box being the only place for 4 to go, and then you can put 7 in the bottom right and then 6 middle right.

So in this case I could put the numbers 1 through 81 and I solve it?

is it still a trio tho if the 8 can be in the middle big box? i thought that only works when the numbers of the trio cant go anywhere else than in the 3 trio boxes

wat

Yes, you can eliminate the is going in the middle box

The 8s

The answer is 3

why the fuck do you put so many pencil marks in per box

3.jpg

I used them when practicing expert mode. Makes it real easy
OP is a brainlet

1-9. Every row and ever column has the numbers 1-9. Only used once in every direction. The bug numbers are the answers for that square. The little numbers are which ones are possible for that square. As a number gets put in, it gets eliminated as a possible answer for every other square in the line or column.

When the puzzles complete you will have a full box, and no number is repeated in a single line.

If what you said was correct then there would be a trio in the bottom left box as well but it's not the case.

Attached: 4.jpg (1080x1080, 284K)

op how long is this puzzle taking you?

Op just do some trial and error my heads fucking splitting

About 10 minutes to get to this point and now I'm stuck.

at this point you run validity scenarios on cells that have 2 candidates, to find the invalid scenario.

it is possible to have a sudoku game ending in various ways, all legal

I gave the cell in the top right block a value of 3. if you see, in doing that, the 3rd column no longer has a candidate of 2 in any of the cells, so it follows that the cell cannot be a 3. if it's not a 3, then it's a 2.

the rest goes on as usual

Attached: where would you put 2.jpg (1173x1194, 210K)

Quality response

That is one based user right here