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March 23 classic difficulty

 
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jrm47
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PostPosted: Fri Mar 24, 2006 7:02 pm    Post subject: March 23 classic difficulty Reply with quote

Ordinarily, given enough time and trial-and-error(threads), I can get through the very hard puzzles, but the March 23rd is giving me a headache.
So far: R5C4(6), R5C5(9), R5C6(1), and R9C5(3), and THAT's IT!. I need one little break-through. Not the solution, but some logic that'll get me one more number. Any help greatly appreciated.
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PostPosted: Fri Mar 24, 2006 8:08 pm    Post subject: Reply with quote

Have a look at R9C9 and R7C6, you should find that there is only one possibility for each of these cells.

Good luck from there.
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windybaer



Joined: 12 Jan 2006
Posts: 1

PostPosted: Sat Mar 25, 2006 9:09 pm    Post subject: Reply with quote

R6C5 is next, but now I'm stuck too.
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Sun Mar 26, 2006 12:48 am    Post subject: There's a hidden pair in r7c1/r8c3 Reply with quote

If you look closely at row 9 and at column 2 you'll see the pair of values {1, 9} lying outside of the bottom left 3x3 box. The way these intersect you can then notice that the pair {1, 9} must occupy the two cells r7c1 & r8c3 -- there's no other way to fit those values in that 3x3 box.

Now there's only one cell left (r7c2) that can hold a "7". And since the only possibilities for r7c9 are {1, 9} you can make more progress in row 7. dcb
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PostPosted: Sun Mar 26, 2006 1:11 pm    Post subject: Reply with quote

OK that was good David
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