September 14th is Cream-Filled Doughnut Day.

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

StevenPatzer wrote:

September 14th is Cream-Filled Doughnut Day.

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

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!

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:

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):

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

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?

I solve it.