Author |
Message |
Cndeariso
Member
06-28-2004
| Saturday, October 28, 2006 - 6:57 pm
julieboo, i've never been stumped this badly before either. are you sure everything else is correct? would you mind posting the starting set and let me see how far i get with it? thanks. i hope i can help you.
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 7:05 pm
CND, the starting set is in my 6:40 post (though you may have seen it by the time I post this...) I am SOOOO glad I am not the only one stumped!
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Cndeariso
Member
06-28-2004
| Saturday, October 28, 2006 - 7:06 pm
working on it right now.
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 7:08 pm
Mame, thanks for sending that flare. And good luck CND. Let me know if/where I've gone wrong... Hopefully you can get at least one or two more numbers...
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Cndeariso
Member
06-28-2004
| Saturday, October 28, 2006 - 7:24 pm
o.k. i cheated and went to an online solver. but, i can tell you this much, julieboo, all that you have so far is correct. you have NO mistakes. now, if you want i can give you one number but i can't tell you how the solver came up with it. let me know.
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Weinermr
Member
08-18-2001
| Saturday, October 28, 2006 - 7:28 pm
Julieboo, I guess Cndeariso beat me to it, but I have a few more numbers for you too. I'm working the puzzle myself.
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Jimmer
Moderator
08-30-2000
| Saturday, October 28, 2006 - 7:34 pm
How does this work anyway?
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 7:35 pm
hmmm, can ya give me any number and explain it???
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 7:36 pm
And thanks!!! Jimmer, each row, each column and each square of 9 have to have 1 thru 9 in each. No repeats. Does that make enough sense? There is no adding involved.
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Cndeariso
Member
06-28-2004
| Saturday, October 28, 2006 - 7:37 pm
Spoiler Click below to view spoiler | according to the logic solver i'm using the next number to be solved is the number 1 (one) in column seven, row E. | but, i'm not sure how it got there. it did a whole series by putting every possible number in each block and used a process of elimination. i really don't know how it got there. maybe someone else can help you with that part.
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 7:37 pm
Jimmer, maybe this can explain better than I can... http://www.sudoku.com/howtosolve.htm
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Jimmer
Moderator
08-30-2000
| Saturday, October 28, 2006 - 7:40 pm
LOL - Thanks Julie. I googled after I asked, and found something similar. It looks like fun.
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 7:41 pm
Thanks CND, I will start with that spoiler if I can't figure anything else out.
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 8:17 pm
Well, I used that "1". Can you tell me if I am right so far?

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Kitt
Member
09-06-2000
| Saturday, October 28, 2006 - 8:23 pm
There's a 1 in the middle row, seventh square along (1st column of the third box. You know the middle row of the very middle box contains 3, 4 and 5 (even though you don't know the order), there's a 2 and a 6 already in that row, so working across the row, squares in columns 2 & 3 & 7 & 8, must contain the numbers 1, 7, 8 and 9 in some order. If you look at column 7 in that middle row, you can't have a 7 or 8 in there, because they're already in the column, you can't have 9 because it's already in the 3by3 box the square it's in, so it must contain the number 1. That's hard to explain, but hope it helped! (ETA: after that you can immediately fill in all the 1s)
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 8:27 pm
Crap, ignore my 8:16 post. Is this one right?

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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 8:27 pm
Kitt, thanks for that. It did help. Now can you tell if I am right so far? I think I am stumped again...
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Kitt
Member
09-06-2000
| Saturday, October 28, 2006 - 8:38 pm
Your 8.26 post is correct, but so is your 8.16 one.
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 8:49 pm
Can anyone explain any one additional number after the 8:26 post? (Hope that makes sense.)
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Kitt
Member
09-06-2000
| Saturday, October 28, 2006 - 8:53 pm
I'll try, give me a sec.. (have to start again from your numbers!)
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Cndeariso
Member
06-28-2004
| Saturday, October 28, 2006 - 9:01 pm
again, i don't know the logic but:
Spoiler Click below to view spoiler | put a 5 (five) in column #2 row F |
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Kitt
Member
09-06-2000
| Saturday, October 28, 2006 - 9:07 pm
OK, this is a bit long winded.. Starting with your 8.26 box: Look at the very middle row again. The middle three in that row are some order of 3, 4 & 5. By eliminating the others in the columns or boxes, square 2 in that row must be 7,8or9, square 3 in the row must be 7or9 and square 8 must be 7or8. Now look at the third column. Those four empty squares must contain the number 3, 5, 7 and 9. In row 7, the square in column 3 must be 7 or 9 (3&5 are already in the row). In row 5, the square in column 3 must be 7 or 9 (see above). That means column 3 rows 1 and 4 must contain the numbers 3 and 5, and col 3 row 1 can't contain 3 as it's already in the row, so col 3 row 1 must be the number 5, and col3 row 4 must be the number 3.
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Chaplin
Member
01-08-2006
| Saturday, October 28, 2006 - 9:27 pm
I solved it and so far you are correct..... Column #1 Row G is a 9
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 9:28 pm
wait...
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Julieboo
Member
02-05-2002
| Saturday, October 28, 2006 - 9:30 pm
why can't col 3/row 1 have a 9?
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