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Let a + b = cNote that 5a - 4a = aNote that 5a - (2 + 2)a = a
Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)cRearranging,
5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c
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
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.
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"
Why doesn't white start off with Qe3?
The fallacy is that if a+b=c, then a+b-c=0, so when you write
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
I'm late today
@dsf001 I see now Kg5 is big blunder. Many Thx.