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Stuck with this for long time now, I have tried all the techniques yhat I know enter image description here

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I will use the notation RXCY, for the cell in row X and column Y, with top left R1C1.

Here's one fill-in you can make

R3C5 cannot be a 9. It must be a 2

Explanation:

If R3C5 is a 9, then R2C5 must be an 8, R5C5 must be a 1, and R5C6 must be a a 7. Deep breath. Now R2C6 must be a 6, R2C4 must be a 7, and R2C7 must be a 1. But wait! Now there is a contradiction. Due to the 1 in the top-right 3-by-3, R3C7 must be a 6. However, due to the 9 in R3C5 (which was the beginning assumption), R3C9 must also be a 6. Since both can't be 6s, the beginning assumption is wrong.

Picture:

depiction of above explanation

Hopefully this gets you un-stuck. If you need more update the question.

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  • $\begingroup$ did you pick R3C5 with any rule.or it was random? $\endgroup$ – Amin Jun 3 at 21:24
  • $\begingroup$ I saw that C5 had a nice set-up where trying any number would automatically give you the column, and I like the number 9. So I tried that first. $\endgroup$ – bobble Jun 3 at 21:27
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I think I have a significantly simpler argument than the current answer: if R8C5 is a 2 then R9C6 is not 2, and also then R3C5 is 9, then R3C2 is the only 2 in its row, then R9C2 is the only 1 in its column, then R9C6 is also not 1, leaving R9C6 with no possibilities, therefore R8C5 is not a 2, therefore R8C5 is an 1.

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