Okay, I think I've got it...there's no math here, just deduction
There are two possible solutions starting with three balls per side.
First solution
1. Start with three balls per side. If these are even, then...
2. Weigh one of the remaining balls against any of the control balls
on the scale. If they are even, then...
3. Weigh the remaining ball against the control ball
to find if it is heavier or lighter---solved.
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Second solution
Part A
1. Start with three balls per side. If these are odd, then...
2. (a) Remove the three balls from the heavier side and set aside.
(b) Take one of the three balls from the lighter side
and place it on the other side of the scale with one of the
two remaining control balls that were originally unweighed.
If these four balls are even then you can logically deduce that
one the three balls from the heavy side (of the original weigh) is
in fact "heavier"...so...
3. (a) Place any two of the three original heavy balls
on the scale. If one of those two balls is heavier, then that
ball is the "heavy" and "odd" ball.
(b) if the scale shows them even, then the remaining ball
is the "heavy" and "odd" ball.
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Second solution
Part B
1. Start with three balls per side. If these are odd, then...
2. (a) Remove the three balls from the heavier side and set aside.
(b) Take one of the three balls from the lighter side
and place it on the other side of the scale with one of the
two remaining control balls that were originally unweighed.
If these four balls are uneven you can now deduce that you
are searching for a "lighter" ball.
3. Clear the two balls from the heavier side of the scale and
weigh the two balls from the lighter side against each other to
find the "lighter" and "odd" ball.
....whew
That took alot of work. Sad thing is, it really isn't that complicated
an answer. But it sure seemed like it...
Anway, thx for the puzzle. It was good one
edit on 7-3-2014 by rival because: (no reason given)