sharara
Age: 124
Total Posts: 5253
Points: 0
Location:
United Arab Emirates, United Arab Emirates
Ok, here's the riddle and solution, and if you find
anything wrong
There are 12 billiard balls on a pool table that all
look exactly alike. They all have the same weight, except
for one. This oddball (no pun intended) has a weight
that is either slightly greater than or slightly less
than the weight of all the other billiard balls. However,
you do not know which ball this is. Using a balancing
scale and nothing else, your job is to figure out which
one has the differing weight. The weight difference
is small enough that you can not tell just by weighing
them with your hands. The limitation is that you can only
use the balancing scale 3 times. Figure it out.
(Hint: it is much harder than you think!)
sharara
Age: 124
8022 days old here
Total Posts: 5253
Points: 0
Location:
United Arab Emirates, United Arab Emirates
SOLUTION:
----------------------------------------------------------
Divide the 12 balls into 3 groups of 4. Name them group A,
group B, and group C. Take group A and set it on the left
of the scale, and set group B on the right of the scale:
A B
oooo oooo
(1) \----/ \----/
| |
\_____________/
Now if the weight is balanced, then set group A and B aside
and take two balls from group C (which you know must contain
the oddball) and set one on each side of the scale and take
two balls from group B and set one on each side of the scale:
C B C B
o o o o
(2) \----/ \----/
| |
\_____________/
If the weight balances here, then take 'C' ball from the two
that you left aside, place it on the right side of the scale,
and take a ball from group B and place it on the left side of
the scale:
B C
o o
(3) \----/ \----/
| |
\_____________/
If this balances, then the oddball is the last ball left in
group C. If there is an imbalance, then the oddball is the
group C billiard ball currently on the scale.
This is just one possible course of events, so backtrack to
figure 2. If the weight does not balance there, then remove
a 'C' ball from one side and a 'B' ball from the other side.
If it balances now, then the C-ball that you just set aside
is the odd one out, and if it does not balance, then the
C-ball currently on the scale is the odd one out.
Now let's backtrack all the way to figure 1. The other
possibility is that group A and group B will not balance.
In that case, remove three A-balls from the left side,
move 3 B-balls from the right side to the left side,
and add 3 C-balls on the right side of the scale:
BBBA CCCB
oooo oooo
(4) \----/ \----/
| |
\_____________/
If the scale stays in the same tilt, then you know that the
oddball is one of the two that did not change position from
figure 1. The only two that did not change position are
the A-ball on the left side and the B-ball on the right
side, so it has to be one of those. Simply place that
A-ball on the left side and put a C on the right, and if
it balances, then the B-ball is the oddball, but if it
does not balance, then the A-ball is obviously the odd one.
case 1)If the scale balances perfectly, then the oddball must be
in the group of 3 A-balls that you removed from the left
side in going from figure 1 to figure 4.
case 2)If the scale tilts in the opposite direction, then the
oddball must be in the group of 3 B-balls that you moved
from the right side to the left side.
For both cases, you can figure out whether the oddball
is heavier or lighter than the normal weight, so place
one of the three on the left side, the second of the three
on the right side, and leave the last one aside for a
moment. If the two balance, then the oddball is the one
you set aside. If they don't balance, the oddball is the
one that is on the appropriate side of the tilt.
All done!
well thats a good solution tooooo
