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HOLLYDOLL Guest

Posted: Tue Aug 09, 2005 9:47 pm Post subject: JULY 3 


AM STUCK  NEED HELP
THANKS
3 4 95
15 392 48
49 5 3
13  4
962 43 58
4  3
5 37 41
31 854 7
4 2 835 

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MD Guest

Posted: Wed Aug 10, 2005 1:46 am Post subject: 


There are pairs in column 1...2/8 and 6/7
The 7 has to go in Column 5..Row 3 because that is the only place it can go because of the 6/7 pair in row 2.
That should help. 

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

Posted: Wed Aug 10, 2005 5:22 am Post subject: July 3  a more intuitive approach 


Hi, Holly!
I think the obvious next move is to place a "1" in r3c7. There isn't any "1" in column 7, yet, and it can't go anywhere else because of "1"'s in r2c2, r1c4, and r7c3.
This leads immediately to a "7" in r3c5, which implies a "1" in r5c5 (the center square) which in turn implies a "7" in r5c9 (the "7" that someone somewhere referred to, without quite explaining how to get there).
Hope that helps. dcb 

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Hollydoll Guest

Posted: Wed Aug 10, 2005 1:53 pm Post subject: Re: July 3  a more intuitive approach 


David Bryant wrote:  Hi, Holly!
I think the obvious next move is to place a "1" in r3c7. There isn't any "1" in column 7, yet, and it can't go anywhere else because of "1"'s in r2c2, r1c4, and r7c3.
This leads immediately to a "7" in r3c5, which implies a "1" in r5c5 (the center square) which in turn implies a "7" in r5c9 (the "7" that someone somewhere referred to, without quite explaining how to get there).
Hope that helps. dcb 
Thanks  I feel dumb for not seeing the single 1 in r3c7.  

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MD Guest

Posted: Wed Aug 10, 2005 1:56 pm Post subject: 


<<This leads immediately to a "7" in r3c5, which implies a "1" in r5c5 (the center square) which in turn implies a "7" in r5c9 (the "7" that someone somewhere referred to, without quite explaining how to get there). >>
The same logic you used was the same I used to get the 7 in r3c5. I mislabeled the column/row in my original post. Regardless, once you get the 1 and/or the 7, the puzzle is done.
GL
MD 

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

Posted: Wed Aug 10, 2005 3:36 pm Post subject: 


You're right  Holly had the hard part done, and was probably just too tired to notice that she was almost there.
I didn't quite understand what you said about the two pairs in column 1. I mean, I saw the two pairs, but couldn't quite follow your explanation about how that led to a "7" in row three. Could you explain that a bit more clearly?
I'm still learning about this stuff. dcb 

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Guest

Posted: Wed Aug 10, 2005 11:07 pm Post subject: 


help...still can't get 1 in row 3 column 7 

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Kim Guest

Posted: Thu Aug 11, 2005 12:35 am Post subject: 


<<help...still can't get 1 in row 3 column 7>>
If you at the same spot as the first post of this thread, then..
r8c7 cannot be a 1
r4c7 cannot be a 1
r2c7 cannot be a 1...
Therefore, to have a 1 in that column, it has to be in r3c7
Does that help???
Kim 

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Guest

Posted: Thu Aug 11, 2005 1:20 am Post subject: 


Kim...You don't know how happy I am that someone took the time to answer my plea for help! thanks, thanks, thanks!!!!
Actually, I'm not even at the some spot as the first question...here's what's still missing in my puzzle:
r3c3 9
r4c1 1
r5c1 9
my beginner's thinking is that r5c1 9 comes first, but I can't figure out why or how...please teach me.
I love doing these puzzles, but am such a novice that even these easy (for some) puzzles drive me nuts!!!! You're great to have taken the time to try to help me. 

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Guest

Posted: Thu Aug 11, 2005 3:18 am Post subject: 


<<my beginner's thinking is that r5c1 9 comes first, but I can't figure out why or how...please teach me. >>
If I read your post correctly, you have everything in the puzzle except for the 3 listed numbers. Anyway, the 9 in r5c1 comes from the fact...
At first glance without obvious elimination, the 9 can either go in r5c1 or r5c9. Did a little deeper in c69....
There is already a 9 in c6r1.
A 9 cannot go in r7c8, r8c8 or r8c8 because it is already full.
So, a 9 has to be in either r4c8 or r6c8.
Therefore, by elimination, a 9 can only go in r5c1. Thus, you have a fixed pair of (1 7) in r5c5 and r5c9.
Hope that helped..
Kim 

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Guest

Posted: Thu Aug 11, 2005 7:24 am Post subject: 


Kim...
Thanks a lot...got confused with some of your instructions, but I got it!
You are great!
cs 

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junior Guest

Posted: Fri Oct 14, 2005 6:12 pm Post subject: Re: July 3  a more intuitive approach 


[quote="David Bryant"]Hi, Holly!
This leads immediately to a "7" in r3c5, which implies a "1" in r5c5 (the center square) which in turn implies a "7" in r5c9 (the "7" that someone somewhere referred to, without quite explaining how to get there).
Sorry for bringing up that ancient puzzle, but how does "this" lead immediately to a 7 in r3c5? Please, help. 

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

Posted: Sat Oct 15, 2005 12:10 am Post subject: Re: July 3  a more intuitive approach 


junior wrote:  David Bryant wrote:  Hi, Holly!
This leads immediately to a "7" in r3c5, which implies a "1" in r5c5 (the center square) which in turn implies a "7" in r5c9 (the "7" that someone somewhere referred to, without quite explaining how to get there). 
Sorry for bringing up that ancient puzzle, but how does "this" lead immediately to a 7 in r3c5? Please, help. 
I suppose you see why the "1" has to go in r3c7. (It has to go in the top right 3x3 box somewhere, and it can't fit in r1c9, in particular, because that would make it impossible to place a "1" in the middle right 3x3 box).
Now analyze the first column. The only numbers missing are {2, 6, 7, 8}. There are both a "2" and an "8" already placed in row 2 and in row 9, Therefore the pair {6, 7} must fit in r2c1 & r9c1, and the {2, 8} pair must occupy r1c1 & r3c1.
But now there's just one place left to place a "7" in row 3  at r3c5  because of the "7"s at r7c6 & r8c8.
I hope that helps. dcb 

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Guest

Posted: Sat Oct 15, 2005 12:51 pm Post subject: Re: July 3  a more intuitive approach 


David Bryant wrote:  junior wrote:  David Bryant wrote:  Hi, Holly!
This leads immediately to a "7" in r3c5, which implies a "1" in r5c5 (the center square) which in turn implies a "7" in r5c9 (the "7" that someone somewhere referred to, without quite explaining how to get there). 
Sorry for bringing up that ancient puzzle, but how does "this" lead immediately to a 7 in r3c5? Please, help. 
I suppose you see why the "1" has to go in r3c7. (It has to go in the top right 3x3 box somewhere, and it can't fit in r1c9, in particular, because that would make it impossible to place a "1" in the middle right 3x3 box).
Now analyze the first column. The only numbers missing are {2, 6, 7, 8}. There are both a "2" and an "8" already placed in row 2 and in row 9, Therefore the pair {6, 7} must fit in r2c1 & r9c1, and the {2, 8} pair must occupy r1c1 & r3c1.
But now there's just one place left to place a "7" in row 3  at r3c5  because of the "7"s at r7c6 & r8c8.
I hope that helps. dcb 
David, thank you VERY MUCH. It was just in front of me, but I couldn't see it.
Junior 

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