# How do I go ahead in this?

Got this from daily challenge of sudoku app.

• Your sudoku image is a bit misleading because in the lower right quadrant, you don't have any 3's marked there for some reason. So I'm not sure that your marks can be trusted at all. – JS1 Feb 9 at 17:54

It's certainly not an easy one to tackle here to find a number. There are probably easier ways to chip away at the pencil marks (after filling in the missing $$3$$s as per JMP's comment), but...

Labeling:
* the cells as $$(column, row)$$ with $$(1,1)$$ in the top-left; and
* sets as $$\{\cdots\}$$

If $$(7,1)$$ were a $$3$$
then $$(7,8)$$ is a $$9$$
and so $$(6,8)$$ would be a $$7$$
leaving $$(6,1)$$ as an $$8$$ and $$(6,2)$$ as a $$3$$
as such $$(5,3)$$ would be a $$1$$
so one of $$\{(7,3), (8,3), (9,3)\}$$ would be a $$3$$
but If $$(7,1)$$ were a $$3$$ none of $$\{(7,3), (8,3), (9,3)\}$$ could be a $$3$$
Therefore $$(7,1)$$ cannot be a $$3$$ and must be a $$9$$

• Idk how you guys have so good eyes to spot such tricks. I try to use pencil markings and eventually I get lost, similar to what happened in this puzzle – Kashish Maheshwari Feb 9 at 18:47
• This was a "guess and check" rather than spotting a pattern. – Jonathan Allan Feb 9 at 18:50

r5c7 can't be a $$3$$, because then r1c7 and r3c7 are $$6$$ and $$9$$, which forces the remaining 2 cells in the column to be $$5,7$$, which forces a $$6,8$$ in the right column of the lower-right block, which means both r7c8 and r9c8 are $$2$$. Also the is only one option for the $$3$$ in r9.