The Times Killer #146

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The Times Killer #146

Postby carlosr » Sat Mar 11, 2006 10:01 pm

The the first time ever I'm completely stuck on a Times killer puzzle:
http://www.timesonline.co.uk/article/0,,18209-2077590,00.html

This is what I've got so far:
Code: Select all
+---+---+---+
|  1|3  |   |
|   | 7 |   |
|   |   |   |
+---+---+---+
|   |   |   |
| 98|5  |   |
|   |  7|   |
+---+---+---+
|   |   |   |
|   |   |   |
|  9|7  |   |
+---+---+---+

Can anyone provide a hint?
carlosr
 
Posts: 7
Joined: 12 December 2005

Postby emm » Sun Mar 12, 2006 5:09 am

r5c1 = (181-180) =1
r6c3 = 17 + 13 + 12 + (16-7) - 45 = 6

c3: 12/3 = 246 and 15/3 = 357
r6 : 8/2 = 35 -> r4c3=7
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Postby carlosr » Sun Mar 12, 2006 12:34 pm

I'd got the r6c3 but forget to put it in my post; it was the obvious-in-hindsight r5c1 that had me stumped. Thanks for the help.

Is it just me or are these getting harder? #147 also seemed trickier than usual.
carlosr
 
Posts: 7
Joined: 12 December 2005

Postby jimbob » Sun Mar 12, 2006 4:03 pm

I think they are getting harder. The Times has changed its source I think.

In my opinion this is a welcome change. It had got to the stage of filling in the numbers without any real thought involved.
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Postby Crazy Girl » Sun Mar 12, 2006 8:22 pm

carlosr/anybody,

How did you get the two 7's in R2C5 and R6C6, also any tip for saturday's Killer number 147
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Postby jimbob » Sun Mar 12, 2006 9:04 pm

For #147 - there is a discussion in the Pappocom puzzles area of the forum. Yes it's in the wrong place - but somehow that's where it ended up.
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Posts: 47
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Two 7s

Postby carlosr » Sun Mar 12, 2006 10:21 pm

How did I get the 7s?

The sum at r6c4 is 23 so it must be 689. There's a 6 at r6c3 (which I forgot to show) so r6c4 must be 8 or 9.

Adding up the bottom-middle and bottom-right nonets you get r6c4 + r6c6 - r7c5 = 7.

Now consider where the 8 and 9 go in the bottom-middle nonet. One of them goes in r7c4/r8c4 and the other goes in r7c5/r7c6. The 9 can't go in r7c6 (because that only leaves 2 in the 11 sum) so suppose r7c6 is 8. This leaves either 1 or 2 in r6c6. It can't be 1 (you'd have 2 in both r7c5 and r7c7) so suppose r6c6 is 2. This implies r7c5 is 3, implying r8c5/c7c6 is 25, implying r9c5/c9c6 is 14, implying r9c7/r9c8 is 25, implying r7c8/r7c9 is 63. Which clashes with the three in r7c5; so r6c6 isn't 2 and r7c6 isn't 8. Hence r7c5 is 8 or 9, and equal to r6c4, so r6c6=7.

r2c5 + r2c6 + r3c6 = 24 (by summing the top-right and middle-middle nonets), so they must be 789. r2c5/r2c6 can't be 89 (that would leave zero in r2c7) so there must be a 7 in r2c5/r2c6.
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