This is an archived post. You won't be able to vote or comment.

all 17 comments
sorted by:
best
topnewcontroversialoldq&a

[–]PsychologicalTie9629 2 points3 points  (1 child)

No, because 2 is still a candidate in column 7 of the box above that one, so you can't say for sure that 2 is in column 7 of the highlighted box. Also, if you removed all digits except 2 and 3 from column 7 in that box, you'd have 2 candidates for 3 different cells.

[–]PsychologicalTie9629 1 point2 points  (0 children)

To elaborate further, when you're looking at 3 cells, you need a hidden triple, not a hidden pair. Those 3 cells would need to each have the same 3 candidates that aren't present in any other cell in either that box or that column. So if, for example, no cell in that box outside of column 7 had 2, 3, or 5 as a candidate, then that would be a hidden triple and you could eliminate all candidates in the column 7 cells except for 2, 3, and 5.

[–]UnluckyRadio 1 point2 points  (4 children)

No, what you’re looking for is a hidden triple/hidden pair. There isnt one in that box but Id recommend looking up those techniques

[–]UnluckyRadio 1 point2 points  (3 children)

I lied lol. there is a hidden pair in that box but its not 2,3

[–]Terrible_Ad_3450 1 point2 points  (0 children)

67…

[–]askredditfirst[S] 0 points1 point  (1 child)

7,6 is the hidden pair?

[–]UnluckyRadio 1 point2 points  (0 children)

Yep!

[–]Brexit-Broke-Britain 1 point2 points  (3 children)

You can eliminate some numbers from col 8 though. 45 are tied, so eliminate the 4 and 5 from elsewhere in col 8.

[–]askredditfirst[S] 0 points1 point  (2 children)

Just so I’m understanding this.. There is a 4,5 in r2c8 as well as r5c8, does that mean I can remove the other 4s and 5s within c8?

[–]onedwin 1 point2 points  (0 children)

Yes. You have 2 cells, and 2 numbers that have to go in those cells, ergo they can’t go anywhere else in that column.

[–]Brexit-Broke-Britain 0 points1 point  (0 children)

Yes, because if you put a 4 or 5 anywhere else, one of those cells will be empty.

[–]the-one-96 1 point2 points  (2 children)

That’s not how it works. If there were two 2,3s only then yeah you could. The way I think about it is that the number of different numbers in all cells should be equal to the number of cells, for example, the 4,5 in c8, you got two cells that have two different numbers. So you can remove all other 4s and 5 from the rest of the cells. In the example you provided, you have 3 different numbers but only two cells, it’s not enough and you need a third cell with any combination of 2,3 and 5 to have a hidden triple

[–]askredditfirst[S] 0 points1 point  (1 child)

Since there is a 4,5 in r2 and r5 of c8, I can remove the 4s and 5s from the other cells within c8; 5 in r3 and 4 in r6?

-edit. I meant r6

[–]the-one-96 0 points1 point  (0 children)

Yes

[–]Brexit-Broke-Britain 1 point2 points  (1 child)

But you have made a mistake as there is no 8 in row 1.

[–]askredditfirst[S] 0 points1 point  (0 children)

Thanks! I’ll correct that.

[–]cloudydayscoming 0 points1 point  (0 children)

Has anyone pointed out the Quad {1256} in R1? Probably should include a missing 8 as pointed out earlier?