A belated counter to BTA's weighing-problem in http://www.sorcerers.net/ubb/Forum7/HTML/000499.html

You have 10 identical purses.
Each containing 10 identical looking coins.
In 9 of the purses are real coins with a mass of 10 grame each,
in the 10th are faked coins with a mass of 11 grame.

How can you determine the purse with the faked coins using a balance scale and optional weights only ONE time?


[This message has been edited by BogiTheWaverer (edited August 23, 2001).]

This could work:

Put 5 bags on each side and adjust with weights to make both sides weigh the same. Then take of one bag from each side, and if the side with the adjustable weights fall down, the bag you just took from the other side has the fake coins.

Don't now if this is to be considered one time, and if the 9 and 11 grames really meant somehting. But nice try, Mollusken!

I can do it if I can add the optional weights until the balance scale is balanced. Is that permissible?

The solution has nothing to do with the specific behaviour of a balance scale.
You can use also a scale at the post office, the "act of weighing" is only for determining a certain mass.

OK here's my solution though I don't know if it satisfies your criteria exactly:

Take 8 coins from Bag 1 and put them on the left side of the balance scale; put the other 2 on the right.

Take all the coins from Bag 2 and put them on the left.

Take 6 coins from Bag 3 and put them on the left; put the other 4 on the right.

Take 3 coins from Bag 4 and put them on the left; put the other 7 on the right.

Take 1 coin from Bag 5 and put it on the left; put the other 9 on the right.

Take 2 coins from Bag 6 and put them on the left; put the other 8 on the right.

Take all the coins from Bag 7 and put them on the right.

Take 4 coins from Bag 8 and put them on the left; put the other 6 on the right.

Take 7 coins from Bag 9 and put them on the left; put the other 3 on the right.

Take 9 coins from Bag 10 and put them on the left; put the other 1 on the right.

Now!

Bag 1 had the fake coins if the left side is 6g heavier than the right.

Bag 2 had the fake coins if the left side is 10g heavier than the right.

Bag 3 had the fake coins if the left side is 2g heavier than the right.

Bag 4 had the fake coins if the right side is 4g heavier than the left.

Bag 5 had the fake coins if the right side is 8g heavier than the left.

Bag 6 had the fake coins if the right side is 6g heavier than the left.

Bag 7 had the fake coins if the right side is 10g heavier than the left.

Bag 8 had the fake coins if the right side is 2g heavier than the left.

Bag 9 had the fake coins if the left side is 4g heavier than the right.

Bag 10 had the fake coins if the left side is 8g heavier than the right.

[This message has been edited by Blackthorne TA (edited August 23, 2001).]

LOL!!!!!!!!!
And only if you had waited a little longer you would have seen the solution and not wasted so much time! lol

:) But where's the fun in that?

That's like saying "If you'd only read the walkthrough, you'd know the answer to the riddle!" :)

[This message has been edited by Blackthorne TA (edited August 23, 2001).]

BTA, are you sure you aren't a descendant from the Oracle in Delphi? All knowing in all matters? ;)

Not bad BTA :) the legend says that british war experts wasted 10.000 hours in common on the solvation of this problem during WW2 and then they suggested to throw down leaflets with this problem on Germany to paralyse the german scientists.

You made the answer a little bit more complicated than it has to be.
The original solvation is:
Take one coin out of the first purse, two from the second, three form the third,....ten from the tenth.
That means 55 coins with a mass of (550 + n )g, at which n e N and 1 <= n <= 10.
The faked coins are in the purse n.


[This message has been edited by BogiTheWaverer (edited August 24, 2001).]

Ah! That solution is much more elegant; shoulda thought of that myself! :)

Good job BTA I almost like his solution better than the offcial one.

Want another one? Here we go.

In WW1 the german troops bombarded Paris with a big calibered cannon called "Dicke Berta" The bombardment caused comparatively less damage and sometimes it led to queernesses.

Once, on the last day of a war-month a shell hited a graveyard and the lone devastation was done on the grave of a renaissance fighter, decorated with his life-size monument holding a spear.
An evening news of Paris reacted on this accident with a logic-problem which caused a sensation:

If you multiply the date (DD) of the day(of the accident) with the length of the spear(in feet), multiply this with half the period of life of the fighter and multiply with half the years that have passed between his death and the bombardment so you get 451066.

How old became the fighter and in which year was his funeral?


[This message has been edited by BogiTheWaverer (edited August 24, 2001).]

But the best riddle of all:

What's in my pocket?

Who knows the answer?

That's not a fair riddle!
I needs three guesses!

BogiTheWaverer:

1) The riddle is not mathematically correct, because you should have said, that all the figures used are Natural numbers.

2) The riddle itself is easy:
- the bombardment took place on February 29, 1918.
- the reneissance fighter was born in 1492 and died at the age of 22 (in 1514).
- his monument depicted him holding a 7-foot long spear.
:D

To those who wonder - the riddle is the most simple case of primal decomposition. I didn't even have to use any programming, the calculator was enough. In the more difficult cases you haveto write some small program that will do the dividing for you. In the most difficult case primal decomposition is used in encryption by banks and special services.

[This message has been edited by Modjahed (edited August 24, 2001).]

the way is right, but in the result is a little error:
1918 is no leap-year and that's why there is no 29th Feb; it's 1916 => he died in 1512