dailysudoku.com

[HOLLYDOLL]

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

[MD]

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.

[David Bryant]

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

[Hollydoll]

[MD]

<<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

[David Bryant]

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

[Guest]

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

[Kim]

<<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

[Guest]

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.

[Guest]

<<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

[Guest]

Kim...

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

You are great!

cs

[junior]

[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.

[David Bryant]

[Guest]