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Fri 22 Jul 2006 (HARD!)

- Code: Select all
`Original Puzzle`

--- 3-9 -7-

--- -2- -61

7-1 --- 3--

4-- 2-- ---

-89 4-5 23-

--- --1 --8

--6 --- 4-5

89- -7- ---

-1- 6-2 ---

Solution So Far

--- 319 -7-

9-- -2- -61

7-1 --6 3--

4-- 2-- ---

189 465 237

--- --1 -48

-76 --- 4-5

89- -7- ---

-1- 6-2 7--

Pencilmarks So Far

256 2456 2458 58 458 48 58 258 24

35 345 3458 578 458 478 58 58 4

25 245 2458 58 458 48 589 2589 249

356 356 357 789 389 378 1569 159 69

- - - - - - - - -

2356 2356 2357 79 39 37 569 59 69

23 23 23 189 389 38 189 1289 239

235 2345 2345 15 345 34 16 12 236

35 345 345 589 34589 348 89 89 39

Well I solved it using the following logic, but damn!, isn't there an easier way?

Mac

Solution:

Because of the pencilmark 58's at r1c7 and r2c7, one must be a 5 and the other an 8. Thus the pencilmark 2589 at r3c8 becomes 29. r3c8 must contain a 2 or a 9.

- Code: Select all
`Looking at the pencilmarks in box 9:`

4* 1289 5*

16 12 236

7* 89 39

If 2 then

4* 89 5*

6 1 2

7* 89 3

If 9 then

4* 12 5*

6 12 3

7* 8 9

Hence there is an 8 in c8 in box 9.

Therefore r3c8=9