S.O.S. Mathematics CyberBoard

[dejavu34]

You have a balance scale and X balls. All of these balls are identical except two, which are heavier than the others by the same amount. You are only allowed to weigh a ball (or balls) against another ball (or balls).

If maximum six weighings will be enough for finding the odd balls, what can be the maximum value for X?

[Denis]

1=3, 2=4, 3=7, 4=10, 5=19, 6=28 (yours), 7=55, 8=82, 9=163 .....

w=weighings, n=number of balls
if w=even : n = 3^(w/2) + 1
if w= odd : n = 3^[(w - 1) / 2 + 1

When n=odd, the 1st weighing is 2 groups of (n-1)/2, leaving 1 aside:
this forces, in the scale=even case, to have only 1 possibility: a Heavy on both sides.

When working with w = even, I had convinced myself that this should be the case again,
since, at 1st weighing, an even number of balls gives 2 possible results instead of 1.

But I was getting nowhere with w=even: someone told me to use n=even,
and weigh instead 1/3 against 1/3 (leaving an extra one aside).

As example, in your 6 weighings case, 1st weighing is 9:9 leaving 10 aside.
Of course, for all w=even cases, then (n-1) must be a multiple of 3.
IF you ask me to type out the steps showing the 6 weighings taking care of
28 balls, the answer is: NO

By the way, the person who helped me out was not commutative
_________________
I upped my standards, so up yours !

[dejavu34]

I know the answer is 30, but I don't know the solution for now.

[Denis]

[Denis]

Hmmm....perhaps 30 IS possible in 6 weighings...

Can you do 12 balls in 4 weighings?
It can be done: just did it with some help.

So we need to reduce the 30 balls to 12 in the 1st 2 weighings...
haven't tried that yet: TRY IT!
_________________
I upped my standards, so up yours !

[Denis]

[Opalg]

[Denis]

Yes, Opalg, but all those concern one ball only: much easier.
I don't think this one (2 heavier balls) is anywhere on the internet.
I had never seen it before.
_________________
I upped my standards, so up yours !

[dejavu34]

[Denis]

[dejavu34]

[Denis]

Is he/she still your friend ?
_________________
I upped my standards, so up yours !

[dejavu34]

Yes, she is my old friend from the school....
Actually, she is very kind person.

I can't change my relationship with her for this event....

But, of course, I will say her that this is not good behaviour.

Thanks.

[helmut]

This is a question from a math competition. Topic has been closed. User dejavu34 has been banned.
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The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low, and achieving our mark. - Michelangelo Buonarroti