Hard Sudoku: hint for next number

Question

I tried a hard sudoku and I can't figure out what the next number is. I tried some online tools without any success (I don't want to have the final solution without any explanation).

I would be very pleased if anyone could give me a hint.

Answer 1

Start by filling in candidates. You should fill them all in, of course, but I'm just keeping things to what's relevant immediately.

Now buckle in, we've got to do a bit of work to get even one number filled in!

Step 1: neither R1,C7 nor R3,C7 can be a 3

R7,C7 can be a 3 or a 7.
- making it a 3 rules out 3 in R1,C7 or R3,C7 as the 3 is taken for C7.
- making it a 7 rules out 7 in R5,C7, so R6,C9 must be the 7 for that block,
  ... which means R1,C9 must be the 3 for C9,
  ... which means 3 is taken for the top-right block, ruling out 3 in R1,C7 or R3,C7.
As both possibilities lead to the same conclusion we know R1,C7 and R3,C7 are not 3.

Step 2: R1,C9 cannot be a 9

In R8, the 6 can be in R8,C1 or R8,C7.
- putting it in R8,C1 means R8,C7 is not a 6, leaving only R9,C8 to be 6 in bottom-right block.
  ... this in turn leaves only R9,C9 to be 9 in that block.
  ... and this in turn means the 9 for C9 is taken, so R1,C9 cannot be 9.
- putting the 6 in R8,C7 rules out 6 in R1,C7 and R3,C7 as the 6 is taken for C7.
  ... so R3,C7 must be 5.
  ... so R1,C7 must be 9, so R1,C9 cannot be.
As both possibilities lead to the same conclusion we know R1,C9 is not 9.

Step 3: R1,C4 cannot be at 9

In C7, the 9 can be in R1,C7 or R4,C7.
- putting it in R1,C7 rules out 9 in R1,C4 as the 9 is taken for R1.
- putting it in R4,C7 rules out 9 in R4,C5 as the 9 is taken for R4.
  ... so R6,C4 must be the 9 for the middle block.
  ... which rules out 9 in R1,C4 as the 9 is taken for C4.
As both possibilities lead to the same conclusion we know R1,C4 is not 9.

Step 4: neither R3,C4 nor R2,C8 can be a 5

R3,C7 can be a 5 or a 6.
- making it a 5 rules out 5 in R3,C4 as the 5 is taken for R3,
  and rules out 5 in R2,C8 as the 5 is taken for the middle-right block.
- making it a 6 rules out 6 in R3,C2 as the 6 is taken for R3.
  and rules out 6 in R8,C7 as the 6 is taken for C7.
  ... which means R8,C1 must be the 6 for R8,
  ... which rules out 6 for R1,C1, as the 6 is taken for C1,
  ... which means R2,C2 must be the 6 for the top-left block (we already ruled out 6 in R3,C2),
  ... which rules out 6 for R2,C6 as the 6 is taken for R2,
  ... making R2,C6 a 5,
  ... which rules out 5 for R3,C4 as the 5 is taken for the top-middle block,
  ... and also rules out 5 for R2,C8 as the 5 is taken for R2.
As both possibilities lead to the same conclusion we know R3,C4 and R2,C8 are not 5.

Step 5: R1,C9 cannot be a 5.

This is rule-out by candidate exhaustion at R7,C3:
5 at R1,C9 means: R3,C7 is 6; R1,C7 is 9; R2,C8 is 4; R3,C8 is 3;
      R8,C9 is 7; R9,C9 is 9; R7,C7 is 3; R7,C8 is 1;
      ... R7,C3 cannot be 1 as the 1 is taken for R7;
      R8,C7 is 5; R9,C8 is 6;
      ... R9,C1 and R9,C2 cannot be 6 as the 6 is taken for R9;
      R8,C1 is 6; R1,C1 is 1; R1,C9 is 5; R1,C3 is 4;
      ... R7,C3 cannot be 4 as the 4 is taken for C3;
      ... R6,C9 is 3; R5,C7 is 7; R5,C1 is 2; R7,C1 is 7;
      ... R7,C3 cannot be 7 as the 7 is taken for R7.

This leaves no candidates for R7,C3, so R1,C9 cannot be 5.

Finally(!) after all that, we can conclude:
R1,C9 must be 3.

Yes, that's a long way to go to get there; if there are any less painful methods to do it than to start exhaustively eliminating candidates in this sort of way, I'm not aware of any.

Answer 2

This one is hard indeed!

The bottom right cell must be either $5$ or $9$. I tried

putting $5$ there and reached a contradiction, but only after a looooong time:



At this point, all possible options lead to contradictions. (The last deduction here, the 6s and 1 on the left, came after assuming the opposite and reaching a contradiction; at the final stage of the gif, there are only two options for the top right 3x3 box, and either of them leads to a contradiction.)

Therefore, the bottom right cell must be

$9$.