## 18th March - Fiendish - #95

Post the puzzle or solving technique that's causing you trouble and someone will help
looking in the top right corner, consider the column with a 2 in
The possibilities for the numbers below are:
389
3678
39
389
Then u get to the 5,
138
1679
then the 4 and its the bottom of the column

Consider the consequences of putting anything but a 1 in the cell below 5
Guest

Hi Jane, 'fraid I reached exactly the same position as you yesterday. I've spent another hour or so this morning and still no joy. It looks like being the first one I've failed on, so hopefully Wayne will consider giving a clue???
Guest

### Fiendish Friday

Ric! WHAT? I've considered the consequences of 3 and 8 below 5. Is it me?
Guest

### In New Hampshire

First, a note about time. If you are a Registered user of the forums, you can adjust all written times to be expressed in your own timezone's time. You just need to go to your "Profile" and change the timezone.

Second, I'm in New Hampshire USA at the moment, so it's not even 6.00 am here. What am I doing - up, at this hour!

Third, to the puzzle in question. |Ric| is close to it - although I think he is inviting you to guess a little. It's not necessary to guess. |Ric| is missing one step.

The answer certainly lies in columns 7 and 8. I hope this doesn't put you all at sixes and sevens - or perhaps I do.

- Wayne
Pappocom

Posts: 599
Joined: 05 March 2005

I don't get it either Ric - neither 3 nor 8 cause immediate contradictions. If you chose 3, the row can be (from the top down):

2 (8 or 9) 7 6 (8 or 9) 5 3 1 4

or 8 gives you

2 (3 or 9) 7 (3,6 or 9) (3,6 or 9) 5 8 1

Neither of these results contradict (as far as I can see) any of the other rows, columns or cells. This is drving me nuts! How do either of these prove it must be a 1?
Guest

OK - That helped Wayne. If you chose an 8 in the top right of the last box that forces means the 6 & 7 both have to go in the bottom middle of the top right hand box (in fact the left-middle box tells you it must be the 7). So that original cell must be a 3. Is that right? That really is fiendish! :o)
Guest

well my point is, if you were to put a 3 or an 8 below the 5 then you cannot complete the column

2
* {3,8,9}
* {3,6,7,8}
* {3,9}
* {3,8,9}
5
* {1,3,8}
* {1,6,7,9}
4

if you were to put a 3 in the cell below the 5
you would have to then complete the {3,9} cell with a 9
therefore one of the {3,8,9} cells is completed but the other one has no possible answer

if you were to put a 8 in the cell below 5
you would have 3 cells that can only take {3,9} - clearly a problem

therefore it must be a 1

n.b. numbers in the {} are the possibilites for the square
Guest

Sorry folks, my last post was unreadable gibberish! Note to self - must re-read before posting!

What I meant was...

If you chose an 8 in the top right of the last box you can complete that box, but this means that the right hand column no longer has a place for a 6 or 7. So looking at the top-right box, only the bottom-middle cell can contain either a 6 or a 7 and it can't be both. Therefore that original cell (Top right in the last box), must be a 3, not an 8.
Guest

Hi Ric - this is getting hectic!

The cell you mark as {3,9} can be a 6 too can't it?
Guest

if u look in column 7, the top cell (left of the 2)
this can not be a 6 due to the 6 in row 1

now, row 9, the 6 has to be in the middle block on row 9. (along with a 2 - this is irrelevant though)

therefore the bottom right hand block can only have a 6 in row 8 (row 7 is full)

the only place a 6 can go in column 7 is in the middle right block

so the 6 can't be in row 8 in middle right block.
Guest

This is good stuff! Why can't the 6 in column 7 go at the bottom of the top-right block too? There is no 6 on row 3.
Guest

the only free position on column 7 in the top right block is in row 1

the other 2 positions are filled with a 1 and a 4
Guest

Posts: 312
Joined: 25 November 2005

Sorry - being a numpty. I miss read. That's very neat.

I have to say though, that both my route and yours feel like following a guess and seeing what happens. I've just got it - with out guessing. Try this...

In column seven, the numbers 6 & 7 only appear in rows 3 and 8, therefore those two cells must be 6 or 7, which eliminates the others in the cell so the options now become...

2
* {3,8,9}
* {6,7}
* {3,9}
* {3,8,9}
5
* {1,3,8}
* {6,7}
4

Number 1 is now only on row 7 so that cell must be 1. Hallelujah!
Guest

|Ric| wrote:well my point is, if you were to put a 3 or an 8 below the 5 then you cannot complete the column

2
* {3,8,9}
* {3,6,7,8}
* {3,9}
* {3,8,9}
5
* {1,3,8}
* {1,6,7,9}
4

if you were to put a 3 in the cell below the 5
you would have to then complete the {3,9} cell with a 9
therefore one of the {3,8,9} cells is completed but the other one has no possible answer

if you were to put a 8 in the cell below 5
you would have 3 cells that can only take {3,9} - clearly a problem

therefore it must be a 1

n.b. numbers in the {} are the possibilites for the square

Ric thanks for your tip. I've been stuck on this puzzle & now understand that the 6 in block 6 has to go in column 7 and therefore you can prove that the 1 must go below the 5 because as you say if you put the 3 or 8 below the 5 you can't complete column 8.
Tim

Posts: 18
Joined: 12 March 2005

well spotted IJ!

i must confess i don't know where thinking one step ahead stops and guessing starts

though I had been stuck at this point for ages and it was actually thanks to a friend I got any further. Though I thought i'd atleast give you lot something to go on
Guest

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