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[weeder]

OK..I know I am missing the obvious. Thanks

1B7F|E-8D|53--|----
-2-0|45F1|-79-|AD--
-5--|7263|D-1C|-F--
--4-|0-BC|-8-F|---7

-C--|F8-6|4-73|--0-
76-E|D0-2|9F--|43--
3F8-|54-B|2-E-|7--9
0-5-|37-E|---D|--F1

B0-7|C--4|8---|-A-F
4--1|-6-9|F--7|-85
--F8|---7|--09|2-4-
-9--|83-F|----|--7-

D---|2F30|-5--|-E--
F---|1ED8|---2|--A-
-8E3|-C45|--F-|B-1D
-1--|-B7A|ED-8|F-C3

[David Bryant]

Hi, weeder! Welcome to the forum.

In the position you posted, you need to concentrate on column 9, and on box 15 (bottom row of 4x4 boxes, right of center).

-- The "0" in column 9 can only appear in box 15 -- that is, at r14c9 or at r15c9.

-- This means that you can eliminate "0" as a possibility at r15c10 & r15c12, in particular.

-- This reveals a triplet {6, 9, A} in row 15, leaving "2" as the sole candidate at r15c1. And right after that you'll find a "2" at r16c14 and a "0" at r16c3 ... that should be enough to get you rolling again. dcb

PS There's also a quintet of values {6, 8, B, C, D} lurking in five cells in box 8 (extreme right, second row of 4x4 boxes). That will allow you to identify r5c14 = 5 (sole candidate, after eliminating "B").

[Suppen]

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