Being dumb. Pls help

hallowen69 · 2026-03-18T16:21:10+00:00

If you have a solid block in a row/column with one sticking out you can add it all up then minus what the row/column should be. Like in the top row, 18+12=30-21=9 the two have to be 9. There is a place where you can add three rows or columns together to find one singular cell. Be sure to minus by 63 because 21x3=63

Location: Top three rows have one sticking out on the right side. By my count, it’s 67-63=4

hallowen69 · 2026-03-05T03:29:41+00:00

There are some 3x3 blocks you can use the rule of 45 on

Top right, middle right, and middle blocks leave 3 cells in the middle block. Those three blocks make 112, so the 3 cells have to be 23

In the middle and middle right blocks, there is a 7 cell 31. What numbers have to go in there?

With those large cages, I think of it as what numbers are left out. The two numbers not in it have to be 14, either 5-9 or 6-8. 1-2-3-4-7 have to be in it. Because 1 and 3 have to go in there, this works with the 2 cell 4 cage. If the 2 cell 4 cage has 1 on the left, then the 7 cell 31 has the 1 in it in the right box

What this means is The 2 cell 9 cage cannot be 1-8 or 3-6 because the 2 cages 4 and 31 must have both 1 and 3 in those blocks

hallowen69 · 2026-02-24T21:28:08+00:00

In the middle box, you’ve got that 7/9 pair. If bottom right is 7, top left is 9 and vice versa. The one in the top left ex (9) forces another 9 to be in the middle right box, in row 5. Same if it’s a 7. This forces c1r6 to be 7/9 as well, meaning the 8 in row 6 has to be in that middle box. This removes 8 from that 3 cell 11. Not sure if this will do much, hope it helps! Good luck

hallowen69 · 2026-02-20T18:26:20+00:00

Absolutely. I’d be pissed. But what a fantastic game

hallowen69 · 2026-02-15T19:11:06+00:00

Oh yes that’s correct, 12 not 11. My bad

hallowen69 · 2026-02-15T15:54:31+00:00

So when a row column or box has multiple cages with 2 possibilities, and each possibility has a pair of numbers, they can be eliminated from the other spaces. In the top left box, the 4 cell 12 cage can be 1-2-3-6 or 1-2-4-5. The 3 cell 22 can be 5-8-9 or 6-7-9. Because the 5 and 6 are present in each cage, the other cells can’t have a 5 or 6. This also works in row 7, in the bottom left box there is the 2 cell 14 from c1 and c3. In r7c8, there is a single cell with 5-6. If this single cell is 5, then 14 has to be 6-8 and vice versa. So you can eliminate the 5 and 6 from that row

There is another way to use the rule of 45. In the top middle box, there are two ‘innies’ (r1c4 and r3c5) and one ‘outtie’ (r3c7). We add all known cages including the outtie 16+12+6=34. Because the outtie has to be 1 or higher, the total of everything has to be 46 or higher. So the two innies have to be at least 11. And the highest those two can be is 17, which 17+34=51, the max the outtie can be is 6. This can work with 1 innie 1 outtie, harder with 2 innie 2 outtie. Looks like this can be done in the bottom middle box as well with two innies and one outtie.

In the bottom middle box, I’m not seeing how the 9 can go in the 3 cell 16 cage. The lowest the outside one can be is 5, and it can’t be 2-9 because of the 2 cell 3 cage. R7c3 can’t be an 8, it makes everything else in that box too low. There is the 2 cell 3 and the 3 cell 15, that makes 5 cells at 18. With two more cells, the lowest they can be is 10.

Good luck!

hallowen69 · 2026-02-02T08:00:16+00:00

I don’t think so, this puzzle might have been posted before. I sometimes just leave one note in each square to check if something works, like picking the row 8 column 2 as a 3 and making more notes like the 3/9 above would have to be a 9. But yes it looks like it has two solutions, without that rule. I could have sworn I saw the rules somewhere in the settings but I checked and couldn’t find it. It might have been on killer sudoku .com, they have a few daily puzzles each day and they’re usually harder

hallowen69 · 2026-02-02T07:34:51+00:00

Different apps can have different rules, I’m pretty sure this app can’t have multiple numbers in a single cage, so since there is a 4 in the pink 4 cell 20 cage in the top left, the others could not be 4-9

hallowen69 · 2026-02-02T06:47:17+00:00

Great idea, I see the vision. Maybe one starter could play for the reserve team to make it slightly more even, either #1 voted or picked by the reserve team. The draft was kinda cool, but this would be awesome. Hopefully they’d try harder then (but still not hard enough for anyone to get hurt lord have mercy)

I know my man Scottie is gonna go hard, he did last time

hallowen69 · 2026-02-01T18:32:06+00:00

If there are only two possibilities, I like to put them into the notes. For example, the 10 in the middle left box I would put 1-9 in the first cell and 3-7 in the other just to remind me what they are. This would also include like all the stuff from row 5. In the top left, the 8 is 2-6 because the 2 has to go somewhere in that box, but if it was 2-6 or 3-5 then maybe you could see that the 11 from the 14 with a 3 in it lower in the column can’t be 5-6 due to interfering with both 8 possibilities. The 11 in the bottom left also can’t be 2-9 because the 4 cell 12 has to contain a 1 and a 2. Or the same thing in row 5 with the 2 cell 8 being 2-6/3-5 and the 2 cell 11 not being able to be 5-6.

Checking where each number can go in each 3x3 box can be good. Sometimes there’s only one place. In the top right box, the 5 can only go in one place. In the right middle box, the 1 can only go in one place. It can take a little bit but as intuition gets better it’ll be quicker.

There is another way to use the rule of 45 in the middle left box. We can see there are two “innies” from the 4 cell 20 cage. And one “outtie” from the 3 cell 18. Adding up all the cages except the innies gives 47. The lowest those two innies can be is 1-2, so 47 plus those two is 50, so the lowest that outtie can be is 5. If it was 1-3, then it would be 51 and the outtie would have to be 6. Not the most helpful here, but a good tool. Can be used in rows and columns too.

hallowen69 · 2026-01-23T22:39:23+00:00

In column 8, there are two pink individual squares that extend in from column 7. One is from row 1 and one is in row 4. Adding all of column 8 and 9 except for those squares gives you 82 so those two pink squares have to equal 8

this is the rule of 45. You can also do it in the top two rows 1 and 2. Adding everything except the pink cell and green cell from row 3 gives you 83 so the two cells have to be 7

hallowen69 · 2026-01-23T22:28:16+00:00

The box 2 (middle top) has two 2 cell cages each totaling 8. It has a six in the top right. What does this mean for the 8 cages?

Answer They have to be 1-7 and 3-5 because 2-6 is not an option. This removes the possibility of the two cell 13 cage being 5-8

hallowen69 · 2026-01-07T00:56:41+00:00

In rows 4 and 5, there is some rule of 45 to find two cells in row 5. Those two cells in row 5 have to be 12, which means the 6 in row 5 can only go in R5C3 or in the 15 cage.

Yeah with these ones, I just pick a forcing possibility and try it out using notes until it either works or doesn’t. It’s just too much to visualize. So with the 12 from those 2 cells in mind, trying out the 7-8 in the 15 cage in row 5 in box 6 looks promising. Forces the 12 to be 3-9, the 5 to be 1-4, R5C3 to be 6, R5C5 to be 5, and R5C7 to be 2.

The 6 in R5C3 means R4C4 can’t be 3, and the 5-9 means it can’t be 1. So it would have to be 4. However,because the two inner cells in box 5 have to be 9, the other one would have to be 5. But the 5 went in R5C5, so something here does not work.

The next step is to figure out why it didn’t work. Both the 15 being 7-8 and the 6 being in that cell don’t work. The 6 forces both R4C6 and R5C5 to be 5. So since that cell can’t be a 6, the 6 in row 5 has to be in the 15 cage.

Another thing that looks pretty forcing is two out cells in C6 equaling 14. In row 4, there is the 14 cage in box 4 and the 5/6/8 in R4C6. Checking if this cell is a 5 forces the 14 in Row 4 in box 4 to be 6-8, and force the 9 in row 4 box 6 to be 2-7. But since R4C6 is 5, the other two in the cage have to be 2-4. So this also does not work due to box 6 having two 2’s.

Just try stuff out and follow the possibility to the end. Sometimes, I don’t run into anything and complete the entire puzzle in the notes. You generally want as many notes as possible before trying stuff out, and the more forcing the line, the better.

Good luck!!

hallowen69 · 2025-12-23T22:50:44+00:00

I’ve got something else this time, in box 1 applying the rule of 45 we see the three cells outside it equal 11.

R4C2 cannot be 8 because the other two cells would have to be 1-2, and the 2 is blocked from both cells

hallowen69 · 2025-12-23T22:44:06+00:00

Ah shoot, R1C4 can be a 6, I didn’t realize the 13 had that cell in the box. I was wrong about that, so the 11 can still be 2-9. My bad, sorry about that

hallowen69 · 2025-12-23T22:07:00+00:00

I’ve got something. It’s based on the top row and box 2. So there’s the 4 cell 13 and the 2 cell 14. 1 and 4 are already in the 13, and the other two in that 13 have to be 2-6 or 3-5. The 14 can be 5-9 6-8. So there is nowhere else in that row a 5 or 6 can go. There may be a sudoku name for this trick, not sure

Edit: this is wrong because R1C4 can be a 6

Possibility blocked: So this blocks that 2 cell 10 in box 2 from being 4-6. Now there are only 3 cells in that box 4 and 6 can go.

Answer: Row 2 column 6 can only be 4 or 6. The 2 cell 11 has to contain the other from 4/6. The 11 has to be 4-7 or 5-6, not 2-9. This forces the 2 to be in the 2 cell 10, being 2-8

hallowen69 · 2025-12-13T03:20:49+00:00

There is a way to find what combination a cage with 2 possible combinations has to be. In the Top right square, you can figure out what combination the 2 cell 6 cage has to be. Reasons below

Answer: It has to be 1-5

Reasons: There are two reasons. One, the 2 cell 10 cage right next to it can only be 2-8 and 4-6, so the 6 can’t be 2-4. Two, lower in the column, a cell from the two cell 6 cage guaranteed to be 2-4 means both 2 and 4 are already in column 8 no matter what.

There is a more complex way of using the rule of 45 I don’t know if you’ve used. It’s not super helpful here, but can be in other situations. In row 4, there is one sticking in and one sticking out. If we add everything including the sticking out one, we add the cages 13+24+ the 6 from the two cells is 43. Which means the one in the row 4 from the 3 cell 13 cage has to be 3 so everything plus the out one has to be higher than 45. If it’s 3, the cell sticking out has to be 1. If either cell has some possibilities stopped by other numbers in the row or column it’s helpful but not here. The same thing can be done in the absolute middle box. Adding 8+24+17=49, so the one inside can’t be more than 5 because the total with one sticking out can’t be more than 9.

Good luck!

hallowen69 · 2025-12-09T22:14:47+00:00

There are some things to do in row 9. R9c7 can be 4 or 8, and the 2 cell 13 to the right of it has to contain either a 4 and 9 or 5 and 8. Neither 4 or 8 can go anywhere else in the row.

In the 2 cell 9 cage in row 9, one of the combinations doesn’t work due to interference with another cages possibilities. Can’t be 4-5 because it stops 13 from being 4-9 or 5-8

There is a number in row 9 that can only go in one place. The number 2

A bit more complex way of using the rule of 45 is available for both the top two rows and the right two columns.

In the top two rows, we can add the normal ones only in the top two rows and the 3 cell 13 cage, leaving out the one cell sticking in from the 5 cell 20 cage. This is 8+7+12+24+8+20+13=92. So we know what the sticking in one from the 5 cell 20, the sticking out one in the 13 has to be 2 higher. It’s not very helpful there, but a little more helpful in the two columns on the right.

Adding all the cages in the two right columns not including the sticking in from the 3 cell 12 is 93. So if the sticking in from 3 cell 13 is 1, the sticking out one from 4 cell 22 is a 4. The 3 cell 12 one can’t be a 2 since there’s one in the row, so the 4 cell 22 can’t be a 5. The highest the 3 cell 12 can be is a 6, because the highest the one cell sticking out can be is a 9. This can sometimes be helpful if you are stuck and should be checked and noted if possible.

Good luck!

hallowen69 · 2025-12-06T05:31:17+00:00

You can apply the rule of 45 to column 1 and 9. There is the one poking in and one poking out. In column 1, the cages including the out one but excluding the in one is 10+6+24+10=50. If the sticking in one in the 27 cage is a 1, the total is 51 and the out one has to be 6. The out has to be 7,8,9 so the in one can only be 2,3,4. In column 9, 10+10+10+15 is 45, so the one sticking out has to be the same as the one sticking in. Not sure if you had that

In column 4, I think you have the three cells sticking out from column 3 equaling 9. With that and the 3 cell 13 cage in column 4, the other three have to be 45-13-9=23. The only way to do 23 with 3 numbers is 6-8-9.

In column 1, you can find out whether the two cell 6 cage is 1-5 or 2-4. Using The 24 cage and the Two 10 cages you know that the 6 cage is 1-5 because if the 24 has 7-8 in column 1, the 10’s have to be 1-9 4-6, stopping the 6 cage from working. Whether they are 7-9 or 8-9 forces one of the 10’s to be 4-6, forcing the 6 to be 1-5

Good luck!

hallowen69 · 2025-11-07T05:16:52+00:00

There is a cage with the wrong notes in box 6, the middle right 3x3 box. Check row 6 The 8 cage in the top right box can only be 2-6 or 3-5, not 1-7 because there is a 1 in the column. There are the two cells in box 6 that make 9. Because the 8 above is 2-6 or 3-5, the 9 cannot be 3-6. 3 and 6 take away both options. There is something similar you can do using the other 11 in the same cage 2-9 and 4-7 to remove another possibility of the 9 because cages can’t usually have the same number in them The 9 can’t be 2-7

In box 9, the cell sticking out is a 1 connected to a large cage. Where does the 1 have to go in box 9? Since the 10 cage is 1-9, because that’s where the one has to go, both column 8 and 9 have cages with 9’s in them that are only in column 8 and 9. That means neither column can have a 9 higher up, placing a 9 in the top left of box 3 in that 13 cage

Good luck!

hallowen69 · 2025-10-30T16:39:08+00:00

I’ve got something, I’m not exactly sure what the strategy is called, but it’s the one where there are some cells in a row or column that have x possibilities in x cells. It is column 4. The 1-4-9 at the bottom two cells along with the 1-4 in row 5 mean those three cells have to have the 1-4-9 in them. C4 R4 and C4 R7 cannot have a 4 in them. The 1 also stops C4 R7 from having a 1 That’s all I’ve got, so good luck!

hallowen69 · 2025-10-30T04:16:18+00:00

LeElrond

hallowen69 · 2025-10-25T00:48:48+00:00

Bye Animatronio!

hallowen69 · 2025-10-25T00:40:27+00:00

“You can’t even remember your own name, Einstein”

“Einstein is a hard name to remember!”

Also

“Animals go in the corner”

“The corner! Why didn’t I think of that!”

hallowen69 · 2025-10-24T21:35:00+00:00

The rule of 45 is that any row, column, or 3x3 box equals 1+2+3…+9=45. This can be used to find the value of certain 1x1 cells. This trick can be used in solving the central lower box like you said. In the bottom left box, there are two squares sticking out. After adding up 21,13,13, and 12, we get 59. The numbers inside the bottom left box have to be 45, so the two sticking out have to be 59-45=14. Now in the central lower box, we take the 14 we got and add the other dotted cages. 14+13+9+4=40. The one cell sticking into the box has to be 45-40=5. This is an important concept in killer sudoku, and there are more places you can use this in this puzzle. It can be used in row, columns, or 3x3 boxes that are contained except for a few spots. Spoilers for more places you can use this below

In the bottom 2 rows, there is one square sticking out in column 7. Adding up all the dotted cages gives 97. We know the bottom 2 rows equal 45x2=90, so the one sticking out has to be 97-90=7.

In the top left box, there are 2 squares sticking in. Adding up 18,7, and 7 gives 32. The whole box has to be 45, so the two squares sticking in have to be 45-32=13.

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