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 5x5 difficult of 1Apr2019 
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Posted on: Mon Apr 01, 2019 5:37 pm




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Joined: Fri Feb 03, 2012 12:14 am
Post 5x5 difficult of 1Apr2019
This puzzle seems unsolvable to me.

I start by filling in the givens...d1, c5, e5.
Now looking at column d, d2 cannot be zero since that would force c2 to also be zero, d4 cannot be zero since either b4/c4 must be a zero, and d5 cannot be zero since either a5/b5 must be zero.
That leaves only d3 to be a zero.
But in order for that clusters subtraction to equal zero that would force e3 and e4 to be the same number.
Stalemate.

What am I missing? Can anyone help?


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Posted on: Mon Apr 01, 2019 8:00 pm




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Joined: Thu May 12, 2011 11:51 pm
Post Re: 5x5 difficult of 1Apr2019
d3 being 0 does not force e3 and e4 to be the same number


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Posted on: Mon Apr 01, 2019 9:35 pm




Posts: 742
Joined: Fri May 13, 2011 6:51 pm
Post Re: 5x5 difficult of 1Apr2019
nashvillebudd wrote:
This puzzle seems unsolvable to me.

I start by filling in the givens...d1, c5, e5.
Now looking at column d, d2 cannot be zero since that would force c2 to also be zero, d4 cannot be zero since either b4/c4 must be a zero, and d5 cannot be zero since either a5/b5 must be zero.
That leaves only d3 to be a zero.
But in order for that clusters subtraction to equal zero that would force e3 and e4 to be the same number.
Stalemate.

What am I missing? Can anyone help?


You are reasoning wrongly because when you have negative numbers in a puzzle like this not necessarily e3 and e4 must be the same number, What about a -1 and a 1, for instance?: 0 - (-1) - 1 = 0 [smile]


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Posted on: Mon Apr 01, 2019 10:13 pm




Posts: 5
Joined: Fri Feb 03, 2012 12:14 am
Post Re: 5x5 difficult of 1Apr2019
sjs34 / clm ...

Thanks for explaining. I was indeed making a logic error. Mentally I had put parentheses around e3 and e4 - doing their subtraction first and then subtracting that result from 0. I used to know that subtraction is not commutative. Funny how my old brain can get stuck on such a simple thing.


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