9/14/2017 - An Abstract Way To Equalize

Vmcmnm

I solve it.

alain978
StevenPatzer wrote:

September 14th is Cream-Filled Doughnut Day.

https://anydayguide.com/calendar/1212

Miam! I'll take a dozen, please!....

DonRR

HUH ???????

chess_guy3000

cool puzzle

szammie

Nice.

SZ.

ismasmanas

good one

ablejack
fatalphenom wrote:

As a newer player, I find no matter what this ending in a stalemate.

I'm also a newer player, but here there are multiple ways to lose. The idea is to avoid what appears to be a losing position. Congratulations on finding the drawn even position. (unlikely to be stalemated) 
/ I managed (w/ no input errors) the draw as well!  

 

Thorin-I

Wow!

Sred

Nothing about this puzzle is abstract in any way.

Gil-Gandel
dsf001 wrote:
Gil-Gandel wrote:
aHorseWithNoName wrote:

Come to my intelligent forum

2+2=5 can it be true?

Let a + b = c
Note that 5a - 4a = a
Note that 5a - (2 + 2)a = a

Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
Rearranging,

5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

Factorising,

5(a + b - c) = (2 + 2)(a + b - c)
Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

5 = 2 + 2

Hope this helps.

@ Gil-Gandel:

The fallacy is that if a+b=c, then a+b-c=0, so when you write

"Factorising,

5(a + b - c) = (2 + 2)(a + b - c)"

what you are really saying is that 5 x 0 = (2+2) x 0, which is true, but you can't cancel the 0s to leave 5 = 2+2 because 5x0/0 does not equal 5 and (2+2)x0/0 does not equal (2+2).

This took about the same amount of time for me to solve as the chess puzzle.

Correct, 10/10, your turn to clean the whiteboard. happy.png

briansladovich
Sred wrote:

Nothing about this puzzle is abstract in any way.

I had to google abstract too, came up with this - 'abstract as a verb' - 2.extract or remove (something):

alain978
aHorseWithNoName wrote:
vagga2 wrote:
aHorseWithNoName wrote:
dsf001 wrote:
Gil-Gandel wrote:
aHorseWithNoName wrote:

Come to my intelligent forum

2+2=5 can it be true?

Let a + b = c
Note that 5a - 4a = a
Note that 5a - (2 + 2)a = a

Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
Rearranging,

5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

Factorising,

5(a + b - c) = (2 + 2)(a + b - c)
Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

5 = 2 + 2

Hope this helps.

@ Gil-Gandel:

The fallacy is that if a+b=c, then a+b-c=0, so when you write

"Factorising,

5(a + b - c) = (2 + 2)(a + b - c)"

what you are really saying is that 5 x 0 = (2+2) x 0, which is true, but you can't cancel the 0s to leave 5 = 2+2 because 5x0/0 does not equal 5 and (2+2)x0/0 does not equal (2+2).

This took about the same amount of time for me to solve as the chess puzzle.

instead of trying to bring new math rules that don't exist, just come to my forum and learn

 

That is an abstract way to equalise.

ur rating is 719. that is way abstract 

Here's another proof that 2+2=5:

US$2. + US $2. = CAN $5.!

And it's true 

dsf001

It is worth noting that after Qxe6+, the king MUST take the queen or else the queen takes the rook!

UmaLautner

Snazzy!

Areliae
Gil-Gandel wrote:
aHorseWithNoName wrote:

Come to my intelligent forum

2+2=5 can it be true?

Let a + b = c
Note that 5a - 4a = a
Note that 5a - (2 + 2)a = a

Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
Rearranging,

5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

Factorising,

5(a + b - c) = (2 + 2)(a + b - c)
Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

5 = 2 + 2

Hope this helps.

Factorising,

Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant

Less sneaky is 5x=4x

cancel the x and you get 5=4

Good misdirection, I approve.

 

 

Crocomules

not related to today's position...   still no English translation of Andre Cheron;  how is this possible?

DavideZ

wow failed

 

dashaign

средненько)))

briansladovich
dsf001 wrote:

It is worth noting that after Qxe6+, the king MUST take the queen or else the queen takes the rook!

Aye, already noted though - comment #52.  

blunderking2

hmm