The New Sudoku Players' Forum

[Aniedunai]

Hello guys, I seriously can´t solve this one, can you help me pls?
This is how far I´ve gone on it.

Code: Select all

7   4 | . 8 . | . . .
. . 5 | . 2 . | . 4 9
2 . . | . 4  .| 3 . 8
------+-------+------
5 . . | 4 . . | . . .
3 4 2| . . . | 8 6 .
1 . . | 6 . . | . . 4
------+-------+------
9 2 .| . 1  7 | 4 . .
. 5 1| . 9  4 | 2 . .
4 7 3| 2 6 5 | 9 8 1 


[m_b_metcalf]

You need to look closely at box 6 (the one containing 4, 6 & 8).

[Aniedunai]

I´m sorry but I cant see it, I´m legally blind xD, all I have is the pencilmarks 1, 9 and 75 from row five, all the free cells from box six can have a 7 or at least I think so.
I don´t have any sure number to any cell from that box.
If you know tell me how you managed to find that out because I really can´t figure it out, maybe I´m overthinking it

Code: Select all

7   4| . 8  .| .  .   .
. . 5| . 2  .| .  4   9
2 . .| . 4  .| 3  .   8
------+-------+--------------
5 . .| 4 .  .| 1  19  .
3 4 2| . 75 .| 8  6   75
1 . .| 6 .  .| .  9   4
------+-------+--------------
9 2 .| . 1  7| 4  .  .
. 5 1| . 9  4| 2  .  .
4 7 3| 2 6  5| 9  8  1


[jco]

Following m_b_metcalf's hint:

there are 3 cells in that box 6 that hold only two digits in each
(look at the ones that "see" more givens; two cells see the same 7 numbers).

That will give you just one placement in that box, but then
in box 3 that placed number has only two possible cells.

[urhegyi]

Without pencilmarks it's hard to continue.
Now it's solved by singles.

Code: Select all

7  1369 4  |159 8  1369|56  1  2
68 1368 5  |17  2  136 |67  4  9
2  169  69 |57  4  196 | 3  57 8
-----------+-----------+--------
5  689 6789| 4  37  28 | 1  29 3
3  4    2  |19  75  19 | 8  6  75
1  89  789 | 6 357  28 |75  29 4
-----------+-----------+---------
9  2   68  | 38 1   7  | 4  35 56
68 5   1   | 38 9   4  | 2  37 67
4  7   3   | 2  6   5  | 9  8  1


[eleven]

There is a naked pair 57 in r5c9,r6c7 (marked x, no other digits possible)
So r4c7 can't be 7, must be 1 (*).

Code: Select all

 7 . 4 | . 8 . | . . .
 . . 5 | . 2 . | . 4 9
 2 . . | . 4  .| 3 . 8
-------+-------+------
 5 . . | 4 . . | * . .
 3 4 2 | . . . | 8 6 x
 1 . . | 6 . . | x . 4
-------+-------+------
 9 2 . | . 1 7 | 4 . .
 . 5 1 | . 9 4 | 2 . .
 4 7 3 | 2 6 5 | 9 8 1

Alternatively you could have seen the 239-hidden triple in r4c89,r6c8 (x) (can't be in r5c9 or r46c7), leaving only r4c7 for 1.

Code: Select all

 7 . 4 | . 8 . | . . .
 . . 5 | . 2 . | . 4*9
 2 . . | . 4  .|*3 . 8
-------+-------+------
 5 . . | 4 . . | * x x
*3 4*2 | . . . | 8 6 .
 1 . . | 6 . . | . x 4
-------+-------+------
 9 2 . | . 1 7 | 4 . .
 . 5 1 | . 9 4 |*2 . .
 4 7 3 | 2 6 5 |*9 8 1

57 can't be in r3c23 or r3c6 (because of the *-marked 57's)
So you have a hidden pair 57 in r3c48, leaving only r1c8 for 1 in box 3.

Code: Select all

*7 . 4 | . 8 . | . # .
 . .*5 | . 2 . | . 4 9
 2 . . | x 4 . | 3 x 8
-------+-------+------
 5 . . | 4 . . | 1 . .
 3 4 2 | . . . | 8 6 x
 1 . . | 6 . . | x . 4
-------+-------+------
 9 2 . | . 1*7 | 4 . .
 . 5 1 | . 9 4 | 2 . .
 4 7 3 | 2 6*5 | 9 8 1

[Aniedunai]

Wow amazing thank you so much for the explanation , I was able to solve it because of the 57 pair on box 6 you guys told me, and also thank you for a deeper insight of other alternatives, definitely learned more because of it.
(Sorry for the late reply, only got a chance to pick it up today)