December! That means it’s time for CHOCOLATE!
My dear friend Essbee showed me this:
Free chocolate ahoy (and white chocolate, my favourite)! But surely there’s got to be a catch?
Of course there’s a catch. You can’t just rearrange an area to end up with a bigger area - moving chocolate around doesn’t magically produce more chocolate. But count ‘em! There are 24 pieces of chocolate ((I’m carefully not calling them ’squares’)) to begin with, and 25 at the end!
Yeah, but it’s one louder
The problem comes with the bits of chocolate in the middle, along the line where they’ve split it. Those blocks, I’m sad to say, are smaller than all of the others. 25% smaller, in fact - and four pieces each losing 25% of their volume makes the one whole block they’ve put in the pot.
Once you know what you’re looking for, it’s easy to see. The right-hand end of the big upper piece that’s split off is clearly three blocks of chocolate tall. When it ends up shifted left by a piece, it has to drop down a quarter of a piece of chocolate - the height of the adjusted bar is 5.75 choccy blocks tall instead of 6.
$4 \times 5.75 = 23$, which is where the missing block has gone to. If you look at these two stills side by side, you’ll notice that the cut-up one is slightly - but definitely - less tall than the original, even though the top three rows remain the same size.
So, sorry, Essbee: you can’t get free chocolate this way.
(You can, however, get free chocolate by going to one of my Games, Goats and Gold talks on the maths of games shows!)
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