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sorry if this has been posted b4

there are six 1 cent coin, and seven 2 cent coins on a table and you can pick up two coins each turn and replace them with one as follows:-

take 1 and 1, replace by 1

take 2 and 2, replace by 1

take 1 and 2, replace by 2

which will be the last coin left?

what if 1 more 2 cent coin is added to original question?

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sorry if this has been posted b4

there are six 1 cent coin, and seven 2 cent coins on a table and you can pick up two coins each turn and replace them with one as follows:-

take 1 and 1, replace by 1

take 2 and 2, replace by 1

take 1 and 2, replace by 2

which will be the last coin left?

what if 1 more 2 cent coin is added to original question?

With even single cent and odd double cent, you'll always end with a 2c coin.

EO = 2

OO = 2

EE = 1

OE = 1

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molly u really nailed it! i know u know how to do this but i still wud like to see the approach.

I'm working on an actual proof when I have free time. It's pretty straight-forward after you realise that you can reduce coins down to 1 immediately.

The key is actually in the second coin. If it's odd, you're left with 2c (since you'll never be able to get rid of that 2c. If it's even, you'll get rid of all of your 2c and never have a chance to get one back.

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