The first move has to be a queen check because all other moves lead to instant mate. There are only four queen checks. Consider each one. Narrowing it down like this helps a lot.

That white King looks very vulnerable and my Queen can only give check but do I wish to loose the Queen?. It might be a light bulb for me. but not just yet. King to g5 comes to mind as a breather but alas "incorrect"

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.

That white Queen looks very vulnerable and my Queen can only give check but do I wish to loose the Queen?. It might be a light bulb for me. but not just yet. King to g5 comes to mind as a breather but alas "incorrect"

Kg5 Qxg4#. Don't forget which way his pawns are going .

it maybe a draw !!!exsiting

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.

impressively at real

Wonderful concept.

John 8:21

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

I am certainly no expert myself, but a stalemate? I was thinking more in the lines of a draw? But I might be mistaken

I don't get the point of that result

Clearly White could Checkmate Black.

Cool!

Good one

The first move has to be a queen check because all other moves lead to instant mate. There are only four queen checks. Consider each one. Narrowing it down like this helps a lot.

Draw

That white King looks very vulnerable and my Queen can only give check but do I wish to loose the Queen?. It might be a light bulb for me. but not just yet. King to g5 comes to mind as a breather but alas "incorrect"

Why doesn't white start off with Qe3?

NİCE ONE

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.

e3 is covered by the Rook on b3. @marco_lapegna_29

That white Queen looks very vulnerable and my Queen can only give check but do I wish to loose the Queen?. It might be a light bulb for me. but not just yet. King to g5 comes to mind as a breather but alas "incorrect"

Kg5 Qxg4#. Don't forget which way his pawns are going .

A very good puzzles for starters but not for me it only made me do some ramdon moves to do it

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, that's why you need to come to my forum. to learn!