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you still dont win

I tried and I tried until what felt like the last legal move came to me finally was the right move. (I could not find the right move because I was looking for check)

Mighty fine.

I solve it.

September 14th is Cream-Filled Doughnut Day.

https://anydayguide.com/calendar/1212

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

HUH ???????

cool puzzle

Nice.

SZ.

good one

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!

Nothing about this puzzle is abstract in any way.

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.

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):

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

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

Snazzy!

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 redundantLess 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?