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JULY 3

 
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HOLLYDOLL
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PostPosted: Tue Aug 09, 2005 9:47 pm    Post subject: JULY 3 Reply with quote

AM STUCK - NEED HELP

THANKS

-3- -4- 95-
-15 392 -48
-49 5- --3

1-3 --- --4
962 4-3 58-
4-- --- 3--

5-- -37 41-
3-1 854 -7-
--4 -2- 835
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MD
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PostPosted: Wed Aug 10, 2005 1:46 am    Post subject: Reply with quote

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

PostPosted: Wed Aug 10, 2005 5:22 am    Post subject: July 3 -- a more intuitive approach Reply with quote

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 Smile
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Hollydoll
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PostPosted: Wed Aug 10, 2005 1:53 pm    Post subject: Re: July 3 -- a more intuitive approach Reply with quote

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 Smile


Thanks - I feel dumb for not seeing the single 1 in r3c7. - Rolling Eyes
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MD
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PostPosted: Wed Aug 10, 2005 1:56 pm    Post subject: Reply with quote

<<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 mis-labeled 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

PostPosted: Wed Aug 10, 2005 3:36 pm    Post subject: Reply with quote

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 Smile
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PostPosted: Wed Aug 10, 2005 11:07 pm    Post subject: Reply with quote

help...still can't get 1 in row 3 column 7
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Kim
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PostPosted: Thu Aug 11, 2005 12:35 am    Post subject: Reply with quote

<<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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PostPosted: Thu Aug 11, 2005 1:20 am    Post subject: Reply with quote

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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PostPosted: Thu Aug 11, 2005 3:18 am    Post subject: Reply with quote

<<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 c6-9....

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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PostPosted: Thu Aug 11, 2005 7:24 am    Post subject: Reply with quote

Kim...

Thanks a lot...got confused with some of your instructions, but I got it!

You are great!

cs
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junior
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PostPosted: Fri Oct 14, 2005 6:12 pm    Post subject: Re: July 3 -- a more intuitive approach Reply with quote

[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

PostPosted: Sat Oct 15, 2005 12:10 am    Post subject: Re: July 3 -- a more intuitive approach Reply with quote

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 Smile
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PostPosted: Sat Oct 15, 2005 12:51 pm    Post subject: Re: July 3 -- a more intuitive approach Reply with quote

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 Smile


David, thank you VERY MUCH. It was just in front of me, but I couldn't see it.

Junior
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