How smart are you?

Question

You have 13 balls and a balancing scale. All the balls look EXACTLY the same, BUT one weighs just slightly different. YOU DON"T KNOW IF THE BALL WEIGHS MORE OR LESS THAN THE OTHERS. You can only use the balancing scale 3 times. How do you discover the odd ball? What is the formula?

PAY ATTENTION!! I said you do not know if the ball is heavier or lighter, if you knew that, this problem could be solved by a 10 year old.

Answer 1

I am smart enough to know that this problem is a well-known one, and rather than try and solve it myself, I have provided you with the link you require instead (sure, it references 'coins' but the principal is the same):

Answer 2

Put 6 balls on each side.
If the scales balances, then you have the odd ball in your hand. If not, take all the balls from the lighter side and split them up with 3 on each side.
Again take the lighter set of balls.
Now take the 3 remaining balls and put one ball on each side of the scale.
If they balance, the odd ball is in your hand. If not, then the lighter side is the odd ball.

Shoot. I didnt read that you didnt know if its lighter or heavier. thats adds a little bit of extra difficulty. I would try and figure it out but I took the time to read the link explaining the answer......

So, basically use the 12 ball solution and if you break even the 13th ball is the odd man out!

good riddle though.

Answer 3

This problem is typically given using twelve balls. I can't seem to wrap my mind around figuring it with thirteen. I will explain it with twelve in case that's what you meant. I'm interested in seeing the answer to figuring it with thirteen balls.

The problem with using two equal divisions of balls is that you don't know if the odd ball is heavier or lighter, so you can not determine which side of the balance has the odd ball when it is unbalanced.

In the twelve ball scenario, you start by splitting the balls into three equal groups of four balls each.

1st weighing -

Weigh one group of 4 against another group of 4.

IF THE FIRST WEIGHING IS BALANCED:

then the odd ball is in the 3rd group.

Weigh three of the four balls from the third group against three “good” balls from those you already weighed. If they balance, you know that the fourth ball is the odd ball and you can weigh it against a “good” ball to determine if it is light or heavy.

If the second weighing doesn’t balance and the balls from the third group fell, weigh two of the three balls you weighed from the third group. If they balance, then the third ball is the odd ball and is heavier. If they don’t balance, the heaviest one is the odd ball and is heavier.

If the second weighing doesn't balance and the balls from the third group rose, weight two of the three balls you weighed from the third group. If they balance, then the third ball is the odd ball and is lighter. If they don't balance, the lighter one is the odd ball and is lighter.

IF THE FIRST WEIGHING IS NOT BALANCED:

Second weighing:

Separate the groupings labeling them heavy balls from the side that fell, light balls from the side that rose, and good balls from the group that was not weighed.

Place three of the balls from the heavy side with the first ball from the light side in the left pan of the balance. In the right pan, place the fourth heavy ball and three “good” balls from the third grouping.

If the left side falls - the odd ball is one of the three heavies. Weigh the first heavy ball against the second heavy ball. If it balances, then the third heavy ball is the odd ball and is heavy. If they don’t balance, the heavier one is the odd ball.

If the left side rises - the odd ball is either the fourth heavy ball or the first light ball that was used. Weigh the light ball against a “good” ball. If it balances, then the fourth heavy ball is the odd one and is heavy. If it doesn’t balance, then the lighter ball is the odd ball and is lighter.

If it balances - the odd ball is one of the three lighter balls that were not used in this weighing. Weight the second lighter ball against the third lighter ball. If it balances, then the fourth light ball is the odd one and is lighter. If it doesn’t balance, then the lightest ball is the odd one and is lighter.



I glanced back at the link cyberdjinn provided and I see it offers a mathematical problem for a thirteen coin scenario. The only problem I see with it is that you have to be certain of at least one "good" coin for it to work. Betting your odds for the "good" coin still leaves you unsure of the final result unless you can use the balance more than three times to ensure you picked the "good" coin in the first place, so I am still uncertain how to solve the exact scenario you have given with complete certainty.

Answer 4

u hold each ball in ur hands and see if u can sense any difference in weight. if so, put them aside. if not, cut each ball in half to see if the material is the same and if u have to, test each material for mineral composition. if the material is different and u know it, set the ball/s aside. if when u tested the mineral composition and one was different, set that ball aside. Now take the ball u have set aside and make sure u have three if not take some out that re the least likely to be the odd ball. Now with those three balls, use the scale to weigh them. Chances are, one of them will be lighter or heavier than the other.

Answer 5

place 7 on each sid if it balance, you know its the remainning one. If not take the side that falls split those 7 into 3 and 3, once again, if it balances then you identified the heavier one as the one remainder not put on the scale from the seven. If not then take the side that falls and put two of the three, if it balances then it's the one left off. If not then it is the one that the scale falls to.

Answer 6

Hi,put 6 balls in each side of the scale,if they weight the same then the ball in your hand is the odd one.
If they dont egual,that means that the odd ball isnt the 1 in your hand,so just take one from this side and 1 from the other,if they r not equal,just keep taken until they r equal(that dosent count us weighting how many times,caz ur just taking out a balls,not putting a ball back)so when they r equal u have two balls in your hand and 1 of them is the odd one,weight them and them you will find out which is the one!Voila and its done!

Answer 7

wow i would say to use the probability formula to make an educated guess. or Just drop a balls and see which one bounces higher or lower then the others, then i would use the inverse of the Force formula, Force MxA=F, so i would use the scale to measure each Mass and weight of the 2 that bounced higher and lower then the rest to find the mass and then i would drop them again to get the A ^_^ there is an easier way but i wanted to sound smart..............

But you could just drop all the balls in water and measure which ball is more dense then the others or is less dense so i would you the density formula

Answer 8

I think you need to use the scale 4 times. You might get lucky and only have to use it once, but to know for sure it takes 4. If you knew one was heavier or lighter then you could do it in only 3.

Answer 9

Leave one of the balls aside. Take the remaining 12 balls and divide them into three equal groups of 4 balls each. Take group A and weigh with group B. If they weigh differently, take the heavier group and split it in half, weighing each pair against the other pair. This will give you the heavier pair. Take one of each of the heavier pair, one in each hand, and determine the heavier one by weighing them in your hands.

If groups A and B weigh the same, then divide group C in half, weigh those pairs and determine the heavier pair. Take the heavier pair, one in each hand, and determine the heavier one.

If groups A, B and C all weigh the same, the heavier ball is the one we set aside at the beginning.

Answer 10

Its so easy lol

TEST #1
======

place 6 on one side= place 6 on another side

Results:

if it is balance: that means, the 13th ball u didnt use is the heavier ball.

if it is unbalance....take those 6 balls on heavier side

TEST # 2
=======
now we have 6 balls

place 3 on one side= place other 3 on other side

Results:
now one side will be heavier...take the 3 balls on heavy side

TEST #3
======
now u have 3 balls only

place 1 on one side= another on another side

Results:

If it is balance.....the one u didnt put on scale is the heavy ball
If it is unbalance...the one on heavier side is the heavier ball

hope that helps =)

Answer 11

if you can only use the balancing scale three times and you can't visibly see any differences or pick them up or bounce them to determine the weights then it is not possible to figure out the different ball unless you happen to choose the ball in one of the first three tries of using the balancing scale.