Brain Teaser No : 00320
You are given 12 balls and one of them has a weight defect - either heavier or lighter. We don't know which one it is. You are allowed to use the balance pan three times.
Find out which one of the twelve is the defective ball?
Submitted by : Nitin Mittal
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Answers submitted by Users : 1-10 of 15
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*Take balls 1, 2, 3, and 4. Weigh 5, 6, 7, and 8 on the first pan, and 9, 10, 11, and 12 on the second pan. (Hereinafter, "Weigh x against y" means "x in the first pan, y in the second.")
**If there is no imbalance, weigh 1 against 2
***Still none? Weigh 1 against 3.
****Still none? 4 is your ball. Imbalance? 3 is your ball.
***Imbalance? Weigh 1 against 3.
****No imbalance? 2 is your ball. Imbalance? 1 is your ball.
**Imbalance? Remember which way the scales tipped. Weigh 7 and 10 against 8, 9, 11 and 12.
***No imabalance? Put 5 on either scale.
****No imbalance? 6 is your ball. Imbalance? 5 is your ball.
***Tipping same direction? Remove 7, and weigh 10 and 11 against 8, 9, and 12.
****Still tipping same? 12 is your ball.
*****Tipping other direction? 11 is your ball.
******No imbalance? 7 is your ball.
***Tipping other direction? Remove 10 and weigh 7 and 8 against 9, 11, and 12.
****Tipping same direction? 9 is your ball.
*****Tipping other direction? 8 is your ball.
******No imbalance? 10 is your ball.Tim Sanders - 17 Jan, 2002 |
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CORRECTION (there were a few technical errors in the first solution)...
*Take balls 1, 2, 3, and 4. Weigh 5, 6, 7, and 8 on the first pan, and 9, 10, 11, and 12 on the second pan. (Hereinafter, "Weigh x against y" means "x in the first pan, y in the second.")
**If there is no imbalance, weigh 1 against 2
***Still none? Weigh 1 against 3.
****Still none? 4 is your ball. Imbalance? 3 is your ball.
***Imbalance? Weigh 1 against 3.
****No imbalance? 2 is your ball. Imbalance? 1 is your ball.
**Imbalance? Remember which way the scales tipped. Weigh 7, 9, and 10 against 8, 11 and 12.
***No imabalance? Put 5 on either scale.
****No imbalance? 6 is your ball. Imbalance? 5 is your ball.
***Tipping same direction? Remove 7, and weigh 1, 10 and 11 against 8, 9, and 12.
****Still tipping same? 12 is your ball.
*****Tipping other direction? 11 is your ball.
******No imbalance? 7 is your ball.
***Tipping other direction? Remove 10 and weigh 1, 7 and 8 against 9, 11, and 12.
****Tipping same direction? 9 is your ball.
*****Tipping other direction? 8 is your ball.
******No imbalance? 10 is your ball.Tim Sanders - 21 Jan, 2002 |
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*you ramdon divide the balls into two even sets (six balls in each) place one set on once side of the scale and one set on the other. The side that moves down is the one with the defect ball.
* you then take the side that weighted more and you pick and four balls and put two on either side.
*If they are equal then you take the remaining two that you didn't use and place them on the scale and the heavier one is the defect.
* If one side is heavier then the other you take the two that are heavier and place them on either side of the scale and the heavier one is the defect.Elysa Laskin - 02 Feb, 2002 |
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you take the 12 balls and divide them up randomly (6 and 6) and place them on the scale. you then take the side that weighs more and divide that up (3 and 3) and place them on the scale. then take the heavier side and choose two from the three and put one on either side of the scale and set the other aside. if they both weigh the same, the one you left out is the heavier one. if one side weighs more than the other the defective one is ovbious. Sandra Braun - 13 Feb, 2003 |
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For some reason this is my favorite puzzle so a am devoting some time to writing the solution.
The basic idea behind solving this puzzle is this:
The maximum number of balls we can sort out with the last one weighing is three. For example; we have three possibly heavy balls. Put one and one on scale and one to the side and you know which one is too heavy.
3 is the magic number.
So whenever there are three of a kind left (heavy or light) and one try at the scale, I will stop explanation.
You can sort them out.
Well, let's get to it.
Place 4 balls against 4 on scale and 4 to the side.
One of three things happens:
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1) We have balance.
A------scale----B to the side C
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NNNN NNNN DDDD
Defective ball is amongst 4 to the side.
We have 8 normal(N) balls and 4 possibly defective(D).
Now weigh 3 defective against 3 normal
ones leaving 1 defective + 5 normal to the side:
A------scale----B to the side C
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DDD NNN DNNNNN
1.1) Balance.
D to the side is defective and we STILL have to weigh it against a normal one to find out if it is to light or heavy.
1.2) A moves down.
One out of 3 D on left side (A) is too heavy.
1.3) A moves up.
One out of 3 D on left side (A) is too light.
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2) A moves up.
A------scale----B to the side C
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LLLL HHHH NNNN
Now we have 4 possibly light(L) and 4 possibly heavy(H) and 4 normal(N).
Weigh 3 light + 1 heavy against 3 normal + 1 light leaving 3 heavy + 1 normal to the side:
A------scale----B to the side C
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LLLH NNNL HHHN
2.1) A moves up.
One out of 3 L on left side (A) is to light.
2.2) A moves down.
Either H on left OR L on right is defective. Weigh either one against a normal one to find out. For example weigh H against N and if balance, L is the devfective one, being too light. If H goes down, it is the defective one being to heavy.
2.3) Balance.
One out of 3 H to the side is too heavy.
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3) A moves down.
A------scale----B to the side C
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HHHH LLLL NNNN
Solution is basically the same as in nr. (2).
Weigh
LLLH against NNNL and HHHN to the side
or
HHHL against NNNH and LLLN to the side
and apply logic explained above ;o)
Done.
Remarks:
Tim Sanders' solution; I didn't dare try understand.
Elysa Laskin and Sandra Braun both assume the defective ball is too heavy.
When I present this problem to people I usually first ask them to solve it with defect ball being to heavy. Once that is solved I change the defect ball to "unknown"
and ask them to have a go at that.
Enjoy!Sigurdur Palsson - 15 Jul, 2003 |
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First group the 12 balls in to 3 groups of 4 each.
Weigh 2 groups. If the defective ball is present in one of these 2, we will know that.
Otherwise, that ball is present in 3rd group.
In eithercase, take that group, and weigh 2 balls once ,2 balls once.Manojna - 07 Aug, 2003 |
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4 4 4
2 2
1 1Subba Rayudu - 08 Feb, 2004 |
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Explanation
we devide the 12 balls in to 4 pairs of
3 balls each
In first weight we place 3 balls on one side of balance and three on other side
if the balance is equale then place ball out
if not than one pair is less or heavy
so we take the less or heavy one pair
then we place one ball aside and one on one side of balance and other is on other side of balance if both are equale then defected ball is which set place aside that not on the balance and if balance is not equale then the balance is less or heavy and take out that ballmandeep singh - 20 Feb, 2004 |
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divide 12 balls into 3 groups each containing
4 balls. put 4 4 balls on each balance pan. this way by using pan 1 times you will get a set of 4 balls which contains the defected ball. than put 2 2 balls on each pan . this way by using weighing balance we have got a set of two balls which contains the defected ball.
now keep one ball in hand and put the other on left pan of the weighing balance and on the right pan of the balance put one ball from the perfect set which we have earlier rejected.
if the weighing balance shows that both balls are of equal weight than the ball in hand is defected otherwise ball kept on left pan is defected.Sourabh Raheja - 03 Jun, 2004 |
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Group 'em in sets of 4 each (a1-a12)
Weigh in grouped sets...would zero in on the defective piece in 3 weighings..!
W1 W2 W3
a1-a4 a5-a8 a9-12
Zero in on the group,then on the defective weight!priyesh - 11 Jun, 2004 |
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