Just found these two puzzles but I can't think of either of the answers, maybe you'll have bettter luck.

1st Puzzle:

Please, divide twelve to two equal parts, and one part must be equal to seven.

2nd Puzzle

You have 10 marked bags with coins. You are short in time and bags are huge. You know, that in 9 bags all coins are true coins, and in one bag all coins are spurious.
The only way you can separate spurious coins from true one is weigh.
(You know, that true coin weigh is 10g and spurious coin weigh is 11g).

So you have tool... the electonic weighing-machine (with one scale).
And you can weight only one time.

Now explain me, how you will decide in which bag are false(spurious) coins.

Good luck

1st puzzle:
Hmm, so far no idea.

2nd puzzle:

If you know, that the bags are marked, then
take one coin from bag 1, two coins from bag 2 and so on.
With 10 bags you will have sum(1..10) = 55 coins.
If they all were true, they would weigh 550g together.

Put them all together on the scale.
The difference from the measured weight to 550g tells you,
how many spurious ones are on the scale, thus from which
bag they came.

- Frank -

Think in terms of other number systems and the other types of division. It is not too difficult

to solve this it helps to be able to count with roman numbers

Haara, with your comment I immediatly saw the solution. Thank you.
Oscar: is your solution also "Roman" based?

Heh, got it.

of course. I just didn't want to make it too easy.