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SR
Joined: 12 Oct 2005 Posts: 21 Location: Cornwall

Posted: Tue Feb 21, 2006 9:51 pm Post subject: 21 Feb v. hard 


3xx 7xx x5x
59x xxx x67
4x7 9x5 8xx
681 xx4 795
xx3 5xx xx8
x45 1xx 326
xx4 6x9 5xx
x3x x5x x8x
x5x xx3 xx1
This is how far I have got
At this stage the hint given is a 2 in r3c9
Can anyone explain the hint?
Sheila 

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WallyGator Guest

Posted: Tue Feb 21, 2006 10:35 pm Post subject: Feb 21 


read down only as much as you need to:
1) There is a hidden pair in row 8 (in 19__ Zsa Zsa Gabor is born).
2) That pair dictates, along with the pair in row 4 (Michael Jordan's number), that a __ (it takes this many to Tango) must appear in column 6...
3) ...and not in r3c5
4) leaving one last pair in row 3
5) making a unique value in r3c8 

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charlie47 Guest

Posted: Tue Feb 21, 2006 10:59 pm Post subject: feb21 


Agree very tough. After several hours and many false starts, I think I figured out the clue.
Cells r4c4 r4c5 are 2/3.
(1) If r4c5 is 2, then r3c5 cannot be 2; so r3c8 must be 2.
(2) If r4c5 is 3, then there is a chain leading to r3c8 being 2.
First c4r4 = 2, and consequently c4r8 =4
Then r9c8 = 4, r7c8 = 7, r7c8 =3, and finally r3c8 = 2
Maybe some less complicated way of getting there.
Charlie 

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

Posted: Wed Feb 22, 2006 1:42 am Post subject: A hidden pair does the trick 


Sheila wrote:  At this stage the hint given is a 2 in r3c9
Can anyone explain the hint? 
Hi, Sheila! I haven't heard from you in a while. FWIW, I thought this puzzle was tougher than the typical "very hard" from Samgj.
Code:  3xx 7xx x5x
59x xxx x67
4x7 9x5 8xx
681 xx4 795
xx3 5xx xx8
x45 1xx 326
xx4 6x9 5xx
x3x x5x x8x
x5x xx3 xx1 
The short answer is that r3c9 = 2 because you can't fit a "2" anywhere else in row 3. Let me explain.
 There's a hidden pair {2, 8} in r1c3 & r2c3 ... if you haven't spotted that already you can find it by seeing the pair {1, 6} in r1c2 & r3c2. Anyway, the point is that there can't be a "2" at r3c2 because of these hidden pairs.
 There can't be a "2" at r3c8 because of the "2" at r6c8. So it looks as if the only spots for a "2" in row 3 are at r3c5 & r3c9.
Now we have to look at the rest of the puzzle. In particular, we'll concentrate on column 6.
 It's clear that the "2" in row 5 must lie in the middle left 3x3 box. So there can't be a "2" at r5c6  and there can't be a "2" at r6c6 because of the aforementioned "2" at r6c8.
 Now let's examine row 8. There's a hidden pair in this row  it's the pair {1, 7} lying in r8c1 & r8c6. To spot it you have to notice the "1" and the "7" in column 3 and also in column 4  then you have to put that together with the "1" in r9c9 and the "7"s in column 7 and in column 9. The upshot of all this is that neither "1" nor "7" can fit in r8c3, r8c4, r8c7, or r8c9  the only place left for the pair {1, 7} is then at r8c1 & r8c6.
 Now you can see that the "2" in column 6 must lie in the top center 3x3 box. So you can eliminate "2" from all the other empty cells in that box. In particular, r3c5 <> 2, and so we must have r3c9 = 2. dcb 

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SR
Joined: 12 Oct 2005 Posts: 21 Location: Cornwall

Posted: Wed Feb 22, 2006 2:10 pm Post subject: 


Thanks, everyone.
I think I was making it harder than it actually was  when I looked, in the light of your advice (and daytime!), I could see I already had the information to force the 2 in the central box of the top line to be in column 6 (because it couldn't be anywhere else in column 6) so that cleared the 2 from r3c5 and everything else came out fine.
I am working my way through the new book, David, and no problems so far, that's why I haven't been on line for a while.
still addicted
Sheila 

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