I would not worry so much about r8c7.

Look at the total of the given boxes in the bottom centre box. That should tell you something about the remaining three boxes.

r7c6 follows

27 posts
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I would not worry so much about r8c7.

Look at the total of the given boxes in the bottom centre box. That should tell you something about the remaining three boxes.

r7c6 follows

Look at the total of the given boxes in the bottom centre box. That should tell you something about the remaining three boxes.

r7c6 follows

- mattlondon
**Posts:**3**Joined:**29 September 2005

OK - so r7c6+r8c4+r9c4 = 22 = 5+8+9 or 6+7+9

9 cannot go in r7c6 so it must be 5 or 6.

6 could go in r7c8 or r7c9 with corresponding 4. But other candidates are possible IF r8c7 is 6.

Do we need to use "If ... then ..." to work it out (same as T&E in my book!) or is there a 'proper' logical step that I'm missing?

9 cannot go in r7c6 so it must be 5 or 6.

6 could go in r7c8 or r7c9 with corresponding 4. But other candidates are possible IF r8c7 is 6.

Do we need to use "If ... then ..." to work it out (same as T&E in my book!) or is there a 'proper' logical step that I'm missing?

- CathyW
**Posts:**316**Joined:**20 June 2005

4 and 5 are eliminated as candidates from r8c7 and r9c7 which leaves 1 and 3 as candidates in r9c7. This gives a 1 and 3 pair with r1c7. Therefore r6c7 cannot be 1, so it must be 2 giving r7c6 as 5.

Realising then, thanks to other tip, where 6 had to go in the bottom centre box led to the rest of the puzzle being completed. Phew:D

Better get on with some work now

Realising then, thanks to other tip, where 6 had to go in the bottom centre box led to the rest of the puzzle being completed. Phew:D

Better get on with some work now

- CathyW
**Posts:**316**Joined:**20 June 2005

I am curious to know how you can be so sure about r6c7, without having got r8c7 first

My method was to add 10, 5 and 8 in bottom centre and deduct from total 45 leaving 22 which can only be formed in 2 ways: 9,7,6 and 9,8,5.

9,7,6 can be excluded because both the 9 and 7 cannot go in r6c7. Therefore, they have to go in r8c4 and r9c4 which is not possible as either a 9 or 7 must be r3c4.

9,8,5 is the other available group for the three spare boxes in the bottom centre group. Only the 5 can go in r7c6. This then gives you the value for r6c7 without knowing r8c7.

My method was to add 10, 5 and 8 in bottom centre and deduct from total 45 leaving 22 which can only be formed in 2 ways: 9,7,6 and 9,8,5.

9,7,6 can be excluded because both the 9 and 7 cannot go in r6c7. Therefore, they have to go in r8c4 and r9c4 which is not possible as either a 9 or 7 must be r3c4.

9,8,5 is the other available group for the three spare boxes in the bottom centre group. Only the 5 can go in r7c6. This then gives you the value for r6c7 without knowing r8c7.

- mattlondon
**Posts:**3**Joined:**29 September 2005

Interesting re r6c7. Both my colleague here and I independently found that in solving this, columns 1 to 7 were completely filled before a single answer was known for sure in columns 8-9. I guess that this is not therefore the 'ideal' puzzle with a single logical thread from beginning to end.

As Dalek said at the beginning, asatisfying puzzle with more sodoku than arithmetic.

As Dalek said at the beginning, asatisfying puzzle with more sodoku than arithmetic.

- Bigtone53
**Posts:**413**Joined:**19 September 2005

I've just worked out that I've messed this one up. That's one of the frustrating things about Killer puzzles. In normal sudoku you realise you've done it wrong because there's two 6s in a box, or because there's nowhere for this 7 to go or whatever. In Killers it sneaks up on you. Me, for instance, I need to fill the 3-cell 15-cage in box 4, but I only have 2-3-6-8-9 left, and I can't do it. Argh!

- PaulIQ164
**Posts:**533**Joined:**16 July 2005

27 posts
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