# 9/14/2017 - An Abstract Way To Equalize

I solve it.

StevenPatzer wrote:

September 14th is Cream-Filled Doughnut Day.

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

HUH ???????

cool puzzle

Nice.

SZ.

good one

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!

Wow!

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.

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.

Sred wrote:

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

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.

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

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

Snazzy!

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.

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

wow failed

средненько)))

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.

hmm