## times puzzle #118 14th April

All about puzzles in newspapers, magazines, and books

### times puzzle #118 14th April

Can anyone help with this? Is there a logic for when there are 3 numbers to go in 3 spaces eg c2 sq 1 (5,8,6) but there are no double pairs?So as example 6 can go in R1 and R3. 5 can go in r2 and r3 and 8 could go in r1,r2,r3
Thanks
Guest

It would help if you could post a representation of where you're up to in the puzzle...

From there you'll probably get a couple of pointers in the right direction.
shakers

Posts: 93
Joined: 10 March 2005

I wish I could help Kleinman but I am completely stuck on this Thursday's Times fiendish puzzle at the following position.

3*1 | 29* | 54*
2*9 | *46 | *1*
4*7 | *** | 9**
------------------
17* | **5 | **9
935 | *** | *7*
62* | 9** | *51
------------------
812 | *** | *95
593 | 81* | **4
746 | **9 | 1*8

Can anyone help me out please with a small clue?

Many thanks.

Zeb
zebedee

Posts: 26
Joined: 27 March 2005

Zeb,

All the 3s in col 7 are in box 3, so are the 7s...
Guest

sure ij is correct - don't know, hada couple of glasses of wine

but 3 in row r5c2 implies that 3 can't be in r5c9, implies that 3 can't be in c9r1-3 , implies 3 must go c7r4-6, therefore 3 must go in c8r9

if this is wrong apologies
Guest

Posts: 312
Joined: 25 November 2005

Anonymous wrote:sure ij is correct - don't know, hada couple of glasses of wine

but 3 in row r5c2 implies that 3 can't be in r5c9, implies that 3 can't be in c9r1-3 , implies 3 must go c7r4-6, therefore 3 must go in c8r9
Guest

IJ wrote:Zeb,

All the 3s in col 7 are in box 3, so are the 7s...

Hi IJ,

Forgive my ignorance, but I'm not sure I follow what you say. I thought that there couldn't be more than one 3 in one box! The position that I reached suggests that a 3 could go in c7r2, c7r4, c7r6 or c7r8. The 7 in c7 could also go in r7 or r8. I'm still confused........

Guest

Posts: 312
Joined: 25 November 2005

sorry to confuse things but

3 can't go in r5c9, so must go somewrere in c9r-1-3, so must go somewhere in c7 r4-6, so in the end must go in c8r9
Guest

gmc wrote:sorry to confuse things but

3 can't go in r5c9, so must go somewrere in c9r-1-3, so must go somewhere in c7 r4-6, so in the end must go in c8r9

Thanks gmc. I follow the logic that, since the 3 can't go in r5c9 it must go in c9r1-3 but I don't see why this necessarily means that a 3 has to then be placed in c7, r4-6, which means a 3 then goes in c8r9. Couldn't a three go in r4,c8 and then in r7,c7 instead? Apologies for being so dense but I'm new to the Sudoku world.

Regards.

Zeb
zebedee

Posts: 26
Joined: 27 March 2005

Anonymous wrote:
IJ wrote:Zeb,

All the 3s in col 7 are in box 3, so are the 7s...

Hi IJ,

Forgive my ignorance, but I'm not sure I follow what you say. I thought that there couldn't be more than one 3 in one box! The position that I reached suggests that a 3 could go in c7r2, c7r4, c7r6 or c7r8. The 7 in c7 could also go in r7 or r8. I'm still confused........

Sorry, I meant col 9 - this probably clarifies things. I am a muppet sometimes! You can probably see that the possible placements for 3 in that column are all in box 3, and likewise for 7s (because r5c9 can't be a 3 or a 7), which means that r2c7 is an 8...
Guest

gmc wrote:sorry to confuse things but

3 can't go in r5c9, so must go somewrere in c9r-1-3, so must go somewhere in c7 r4-6, so in the end must go in c8r9

I don't follow the leap from c9r1-3 to c7r4-5 - why can't the 3 in box 6 go in r4c8?
Guest

IJ wrote:
Anonymous wrote:
IJ wrote:Zeb,

All the 3s in col 7 are in box 3, so are the 7s...

Hi IJ,

Forgive my ignorance, but I'm not sure I follow what you say. I thought that there couldn't be more than one 3 in one box! The position that I reached suggests that a 3 could go in c7r2, c7r4, c7r6 or c7r8. The 7 in c7 could also go in r7 or r8. I'm still confused........

Sorry, I meant col 9 - this probably clarifies things. I am a muppet sometimes! You can probably see that the possible placements for 3 in that column are all in box 3, and likewise for 7s (because r5c9 can't be a 3 or a 7), which means that r2c7 is an 8...

Thanks IJ. All is now clear. I need to improve my lateral thinking a tad to solve the fiendish puzzles.

Cheers

Zeb
zebedee

Posts: 26
Joined: 27 March 2005

sorry, i was responding from memory (and after a couple of glasses of wine which is always a mistake!)

That was the logic i used at one point so i must have already had something in c8r3 - i'll just have to have another go at it.
Guest