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Weighing IV.

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Weighing IV. - Back to the Water and Weighing Puzzles

One of twelve tennis balls is a bit lighter or heavier (you do not know which) than the others. How would you identify this odd ball if you could use an old two-pan balance scale only 3 times?

You can only balance one set of balls against another, so no reference weights and no weight measurements.

This old topic is locked since it was answered many times. You can check solution in the Spoiler below.

Pls visit New Puzzles section to see always fresh brain teasers.

Weighing IV. - solution

It is enough to use the pair of scales just 3 times. Let’s mark the balls using numbers from 1 to 12 and these special symbols:

x? means I know nothing about ball number x;

x< means that this ball is maybe lighter then the others;

x> means that this ball is maybe heavier then the others;

x. means this ball is “normal”.

At first, I lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.

If there is equilibrium, then the wrong ball is among balls 9-12. I put 1. 2. 3. on the left and 9? 10? 11? on the right pan.

If there is equilibrium, then the wrong ball is number 12 and comparing it with another ball I find out if it is heavier or lighter.

If the left pan is heavier, I know that 12 is normal and 9< 10< 11<. I weigh 9< and 10<.

If they are the same weight, then ball 11 is lighter then all other balls.

If they are not the same weight, then the lighter ball is the one up.

If the right pan is heavier, then 9> 10> and 11> and the procedure is similar to the former text.

If the left pan is heavier, then 1> 2> 3> 4>, 5< 6< 7< 8< and 9. 10. 11. 12. Now I lay on the left pan 1> 2> 3> 5< and on the right pan 4> 9. 10. 11.

If there is equilibrium, then the suspicious balls are 6< 7< and 8<. Identifying the wrong one is similar to the former case of 9< 10< 11<

If the left pan is lighter, then the wrong ball can be 5< or 4>. I compare for instance 1. and 4>. If they weigh the same, then ball 5 is lighter the all the others. Otherwise ball 4 is heavier (is down).

If the left pan is heavier, then all balls are normal except for 1> 2> and 3>. Identifying the wrong ball among 3 balls was described earlier.

One of twelve pool balls is a bit lighter or heavier (you do not know) than the others. At least how many times do you have to use an old pair of scales to identify this ball (and be always 100% sure)?

(a pair of scales = a scale consisting of a lever resting on a fulcrum with weighing pans at each end of the lever equidistant from the fulcrum)

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I have a shorter solution.

Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.

From these balls, put three balls on each side of the scale.

Whichever is lighter, the lighter ball is in that group of three.

From the three balls, put a ball on each side of the scale.

Case1: If the scales balance, the third ball is the lighter ball.

Case2: If one is lighter, then that is the ball...

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I have a shorter solution.

Put six balls on each side of the scale. Whichever is lighter, the lighter ball is in that group of six.

From these balls, put three balls on each side of the scale.

Whichever is lighter, the lighter ball is in that group of three.

From the three balls, put a ball on each side of the scale.

Case1: If the scales balance, the third ball is the lighter ball.

Case2: If one is lighter, then that is the ball...

Just a little detail you missed:

One of twelve pool balls is a bit lighter or heavier (you do not know)

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I have another solution. Use the scale just twice. Place 6 balls in each. One of the scale is down at this point. Keep removing one ball from each scale. At one point the scale will balance. Now we know that one of the two balls is lighter or heavier than the rest. Clear the scale and place one of the 2 balls in one scale and another ball from the remaining group on the other. if they balance, then the other ball is the lighter or heavier one, if not that is the one !

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The very least amount of times you could weigh would be 2. You pick two balls number 1 and number 2, you weigh them and they happen to be different weights. You know that one of these balls is the one you are looking for. You weigh balls number 2 and number 3, if they are equal in weight than it is ball number 1, if they are different in weight than it is ball number 2.

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I think the best way to resolve this problem is by building a deterministic scale array. Every combinations of balls should not be evaluated more then once against another group, then 3 balls will appear once, 3 will appear 3 times and the other 6 balls will be tested 2 times.

A sample array will be as follow:

1 2 3 7 | 8 9 10 4

3 4 5 8 | 9 7 11 6

5 6 1 9 | 7 8 12 2

Then to find any combination, you simply evaluate the 3 balance results:

Ball 1 heavier:

1 2 3 7 > 8 9 10 4

3 4 5 8 = 9 7 11 6

5 6 1 9 > 7 8 12 2

Ball 2 heavier:

1 2 3 7 > 8 9 10 4

3 4 5 8 = 9 7 11 6

5 6 1 9 < 7 8 12 2

Ball 3 heavier:

1 2 3 7 > 8 9 10 4

3 4 5 8 > 9 7 11 6

5 6 1 9 = 7 8 12 2

Ball 7 lighter:

1 2 3 7 < 8 9 10 4

3 4 5 8 > 9 7 11 6

5 6 1 9 > 7 8 12 2

Ball 10 heavier:

1 2 3 7 < 8 9 10 4

3 4 5 8 = 9 7 11 6

5 6 1 9 = 7 8 12 2

Another virtue to this method is that it is easily programmable.

.

/S

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Each weighing has three outcomes:

Left side is [lighter than] [equal to] [heavier than] Right side.

Three weighings can thus discern among [3]x[3]x[3]= 27 cases.

We have only 24 cases:

one of 12 balls is heavier or lighter than the rest.

So we can solve the problem, so long as ...

[1] The first weighing reduces the cases to no more than 9.

[2] The second weighing reduces the cases to no more than 3.

[3] The third weighing then distinguishes among 3 or fewer cases.

First weighing:

Set aside four balls. Why?

Because, if the first weighing balances, we have 8 [fewer than 9] cases:

One of the 4 excluded balls is heavier or lighter.

[1] Weigh 1 2 3 4 5 6 7 8

Outcome 1[a] 1 2 3 4 balances 5 6 7 8.

Since only one ball is odd, 1 2 3 4 5 6 7 8 must all be normal.

We have 8 cases: 9 10 11 or 12 is H or L

[shorthand: H=heavier; L=lighter]

[2] Weigh 1 2 3 9 10 11 [we know 1 2 3 are normal]

If this balances, 9 10 11 are also normal, and 12 is H or L.

[3] Weigh 12 [any other ball]

If 12 rises, then 12L; if 12 falls, then 12H.

If [2] 9 10 11 rises, we have 3 cases: 9L, 10L or 11L

[3] Weigh 9 10.

If [3] balances, then 11L;

If 10 rises, then 10L; If 9 rises, then 9L

If [2] 9 10 11 falls, we have 3 cases: 9H, 10H or 11H

[3] Weigh 9 10.

If [3] balances, then 11H;

If 10 falls, then 10H; If 9 falls, then 9H

end of Outcome 1[a]: balance.

Outcome 1: 1 2 3 4 falls, and 5 6 7 8 rises.

This gives 8 cases: 1H, 2H, 3H, 4H, 5L, 6L, 7L, or 8L

The second weighing in this case becomes tricky.

Remember each of its three outcomes can lead to no

more than 3 cases for the third weighing to resolve.

Again, we exclude a number of balls and involve the others.

We exclude any three of the balls. Why? Because if the

other [included] balls balance, we have exactly 3 cases.

Without loss of generality we exclude balls 1 2 3.

Since that leaves an odd number of balls, 4 5 6 7 8,

we need to use one of the normal balls.

Finally we choose which three to weigh against the others.

And here's the only hard part of this problem.

We must mix some of the possibly light balls with some

of the possibly heavy balls. Otherwise, one of the

outcomes of the second weighing will leave us with more

than 3 cases, and the third weighing will not resolve this.

[2] weigh 4[H] 5[L] 6[L] 1[normal] 7[L] 8[L]

in parentheses I've indicated the POSSIBLE cases

that we have determined:

4[H] means 4 is heavier if it's not normal.

Outcome 2[a]: 4 5 6 balances 1 7 8

These balls are all normal.

We have 3 cases: 1H 2H or 3H.

[3] weigh 1 2

If [3] balances, then 1 and 2 are normal, and 3H

if 2 falls, then 2H

if 1 falls, then 1H

Outcome 2b: 4 5 6 falls, and 1 7 8 rises.

We have 3 cases: 4H 7L or 8L

[3] weigh 7 8

if [3] balances, then 7 and 8 are normal, and 4H

if 7 rises, then 7L

if 8 rises, then 8L

Outcome 2[c]: 4 5 6 rises, and 1 7 8 falls.

We have 2 cases: 5L or 6L.

[3] weigh 1[normal] 5[L]

If 5 rises, then 5L

If balance, then 5 is normal, and 6L

Now we can go back and take the remaining case

Outcome 1[c]: 1 2 3 4 rises, and 5 6 7 8 falls.

to distinguish among the remaining 8 cases:

1L 2L 3L 4L 5H 6H 7H and 8H

exactly as we analyzed Outcome 1b.

Simply substitute H for L and v.v.

Problem solved.

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1 weigh 4 balls on scale - groups a,b & c- if a & b balance the oddball is in c, go to step 3

2 if unbalanced hold a and balance b- 2 on each side- if equal discard

3 balance a- repeat 2 & 3 until you have 1 ball on each side. the oddball and another

4 pick any other ball to balance off the finalists-if equal discard

if your lucky you can confirm the oddball in as few as 4 tries

cai

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i think that this would be easier,

1st weighing - weigh 6 balls on each pan and the oddball is in the uneven one.

2nd weighing - weigh 3 balls in each pan and the oddball is in the uneven one.

3rd weighing - weigh any two balls the oddball is in the un even one. if they are even, the oddball is the remaining.

right?

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I have another solution. Use the scale just twice. Place 6 balls in each. One of the scale is down at this point. Keep removing one ball from each scale.

Each time you remove a ball and look at the scales, it's considered a separate weighing.

The very least amount of times you could weigh would be 2.

Actually, the least amount of times you could weigh to identify which ball is different is once. This, however, is not what the riddle is asking for. It's asking (bolding mine): "At least how many times do you have to use an old pair of scales to identify this ball (and be always 100% sure)?"

1st weighing - weigh 6 balls on each pan and the oddball is in the uneven one.

If you're weighing two groups of balls, there is no oddball group; they are both different from one another. You won't know which group to separate for the second weighing.

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To clarify:

The solution of weighing 6 and 6, 3 and 3, and then finding the oddball doesn't work. When you get one group to be heavier and one group to be lighter, you still don't know whether the oddball is heavier or lighter-- if it's heavier, you want to start weighing the heavy group, if it's lighter, you want to start weighing the lighter 6.

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The answer is three.

Separate the balls in three groups of four and place a letter on them, from A to L.

The known balls will be called "X".

So, here we go:

ABCD EFGH IJKL

The 1st weighing.

Use the first two groups.

There will be three results:

1) ABCD EFGH (I) => Group IJKL contains "The Ball"; ABCD EFGH will be called XXXX XXXX

or

2) ABCD ____ (I) => IJKL will be called XXXX

_______ EFGH

or

3) ____ EFGH (I)

_ ABCD

Obviously, the solution for case 3 is the same for case 2.

Solution for Case 1.

First result:

1) ABCD EFGH (I) => Group IJKL contains "The Ball"

4) IJK XXX (II) => L<>X

5) L _ (III) => L is the lightest ball

___ X

or

6) _X (III) => L is the heaviest ball

__L

Second result:

1) ABCD EFGH (I) => Group IJKL contains "The Ball"

7) IJK ___ (II) => IJK is lighter than the others

_____ XXX

8) I J (III) => K is the lightest ball

or

9) I __ (III) => K is the lightest ball

___ J

or

10) _J (III) => K is the lightest ball

___I

Third result:

1) ABCD EFGH (I) => Group IJKL contains "The Ball"

11) ___ XXX (II) => IJK is heavier than the others

___ IJK

12) Use solutions 8,9 and 10 changing "lightest" for "heaviest".

Solution for Case 2.

First result:

2) ABCD ____ (I) => IJKL will be called XXXX

_______ EFGH

13) AFGH EXXX (II) => BCD is lighter than the others

14) Use solutions 8,9 and 10 to find "The Ball".

Second result:

2) ABCD ____ (I) => IJKL will be called XXXX

_______ EFGH

15) AFGH ____ (II) => A<>E

________ EXXX

16) Use solutions 5 and 6 to find "The Ball".

Third result:

2) ABCD _____ (I) => IJKL will be called XXXX

_______ EFGH

17) _____ EXXX (II) => FGH is heavier than the others

___ AFGH

18) Use solution 12 to find "The Ball".

That's all.

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How about you just put all the balls in one side then balance the other side with weights. Then take each ball and put it back till the “ball side of the scale goes up? :-)

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I have another solution with three steps that follows similar logic to the author/admin:

using the same terminology of x>, x<, x? and x:

the first step is the same, but I will repeat for completeness!

put 1? 2? 3? 4? on the left, 5? 6? 7? 8? on the left,

if there is equilibrium

____ we know that the odd ball is 9?, 10?, 11?, 12?

____ we now put 5, 6, 7 on the left and 9?, 10?, 11? on the right

____ if there is equilibrium

________ we know that 12? is the odd ball, we weigh it against, say 5

________ if it is lighter, then answer is 12<

________ else if it is heavier, then the answer is 12>

____ else if the left is heavier

________ we know that the odd ball is 9<, 10< or 11<

________ we now put 9< on the left and 10< on the right

________ if there is equilibrium then the answer is 11<

________ else if the left is heavier, then the answer is 10<

________ else if the right is heavier then the answer 9<

____ else if the right is heavier

________ (same as previous case with > substituting

________ we know that the odd ball is 9>, 10> or 11>

________ we now put 9> on the left and 10> on the right

________ if there is equilibrium then the answer is 11>

________ else if the left is heavier, then the answer is 9>

________ else if the right is heavier then the answer 10>

________ (this concludes the part that is the same as the author)

else if left is heavier

____ we know that 9,10,11,12 are not odd and that we have 1>, 2>,3>,4> and 5<,6<,7<,8<

____ we now put 1>,2>,5< on the left and 3>,4>,9 on the right

____ if there is equilibrium

________ we know that 1,2,3,4,5 are not odd so we are left with 6<,7<,8<

________ we now put 6< on the left and 7< on the right

________ if there is equilibrium then the answer is 8<

________ if left is heavier then the answer is 6<

________ if the right is heavier then the answer is 7<

____ else if the left is heavier then

________ we know that 3,4,5,6,7,8 are not odd, and we are left with 1>,2>

________ we now put 1> on the left and 2> on the right

________ if the left is heavier then the answer is 1>

________ if the right is heavier then the answer is 2>

____ else if the right is heavier then

________ we know that 1,2,6,7,8 are not odd, so we are left with 3>,4>,5<

________ we now put 3> on the left and 4> on the right

________ if we have equilibrium then the answer is 5<

________ if we left is lighter then the answer is 3>

________ if the right is lighter then the answer is 4>

else if right is heavier

____ the same as previous (wide) condition as left and right in this case are interchangeable (i.e. exchange 1,2,3,4 with 5,6,7,8 and follow same logic!)

I think this is correct, can anyone disqualify this?

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Why Can't I delete my comment.

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Why Can't I delete my comment.

only I can do that

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Weigh 4 balls on left, 4 balls on right, rest left off.

Gives three possibilities.

1) Left side is heavier (down).

2) Left side is lighter (up).

3) Equilibrium.

-

For 2), turn the scale around so we only have two possibilities.

-

For 1), either the left contains a heavy ball or the right contains a light ball, but the rest is definitely normal.

Replace the right side with the remainder. Weigh.

Two possibilities:

A) Left side is still heavier; it contains a heavy ball.

B)Equilibrium, right side contained light ball.

Therefore, we know there is a group or four and whether the ball is heavy or light.

Put two balls on the scale, weigh.

Same three possibilities, as in the beginning.

For 1, right ball is the light ball or left is the heavy ball, as determined. (3 weighs)

For 2, vice versa. (3 weighs)

For 3, weigh the remaining two balls, and you'll get 1) or 2). (4 weighs)

-

For 3) on first weigh, place two balls from the left-off group and weigh.

1. Left ball heavier

2. Right ball heavier, turn it around and make 1)

3. equilibrium.

For 1., replace right ball with a left-off ball and weigh.

A) Left still heavier, left ball is the heavy ball. (3 weighs)

B) Equilibrium. The taken off ball is the light ball. (3 weighs)

For 3., switch the other two and weigh, the proceed as 1. or 2. (4 weighs)

max: 4 weighs.

to osoriocj, 8), 9), 10) maken't sense.

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All the solutions I read were too complex to understand or simplified and didn't sound legit.

Here is my solution:

1: Three groups of Four Balls (A,B,C)

2:

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All the solutions I read were too complex to understand or simplified and didn't sound legit.

Here is my solution, in which, you need 4 tests to find the solution in any case, I wrote it out as simple as possible:

Three groups of Four Balls (1-4, 5-8, 9-12)

Test 1) weigh 1-4 vs 5-8

Test 2) weigh (either 1-4 or 5-8) vs 9-12

with these two tests we will know which group has the odd ball and whether it is heavy or light

Case: 1-4 = 5-8

Therefore: 9-12 has odd ball and test 2 will tell us whether it is heavier or lighter

Case: 1-4 is not equal to 5-8

Therefore: 9-12 is normal, test 2 will show which group has the odd ball (no matter which group you pick to weigh against) and if it is heavy or light

now the group that has the odd one, the balls are a, b, c, and d

Test 3) weigh a,b vs c,d

Test 4) weigh (a or b) vs (c or d)

this will let us know which ball it is

We already know if it is heavy or light, the third test will show which group of two it is in, the last one will show which exact ball it is (if you took the odd ball out, the scales will even out and you will know it was the odd one)

hope this is an easier read than all the others

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While my solution is a max of 4 steps, i think it is much simpler than some of the previously mentioned solutions.

(Answers are denoted with a * )

First you break the balls up into 4 groups of 3: A(1-3), B(4-6), C(7-9), D(10-12)

1: Test A and B

_1[1](A=B): The oddball is in C or D

__2: Test A and C

__2[1](A=C):The oddball is in D.

___3:Test 10 and 11

*__3[1](10=11): The oddball is 12

___3[2](10<>11):10 or 11 is the oddball

____4:Test 10 and 12

*___4[1](10=12): The oddball is 11

*___4[2](10<>12): The oddball is 10

__2[2](A<>C): The oddball is in C

___3: Test 7 and 8

*__3[1](7=8): The oddball is 9

___3[2](7<>8):7 or 8 is the oddball

____4:Test 7 and 9

*___4[1](7=9): The oddball is 8

*___4[2](7<>9): The oddball is 7

_1[2](A<>B): The oddball is in A or B

__2:Test A and C

__2[1](A=C):The oddball is in B

___3:Test 4 and 5

*__3[1](4=5): The oddball is 6

___3[2](4<>5):4 or 5 is the oddball

____4:Test 4 and 6

*___4[1](4=6): The oddball is 5

*___4[2](4<>6): The oddball is 4

__2[2](A<>C):The oddball is in A

___3:Test 1 and 2

*__3[1](1=2): The oddball is 3

___3[2](1<>2):1 or 2 is the oddball

____4:Test 1 and 3

*___4[1](1=3): The oddball is 2

*___4[2](1<>3): The oddball is 1

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My graphical representation of the solution for only 3 weighs... might be a little difficult to read, but see if you can follow.

post-4361-1203718591_thumbjpg

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3 is the minimum number of uses of the scale. My basketball coach in high school told the team that if anyone could solve this puzzle the team would avoid running excessive sprints that week. His version was a tad different, (he wanted the method, and told us we could only use the scale 3 times) but the principles are the same. After solving for three weighings I don't believe it's possible to get enough information with only 2 uses of the scale.

The key to solving this (as "bonanova" shows) is realizing that you get more information out of the unbalanced scale than most people realize at first. Say you put to piles of four on either side of the scale. If it is uneven, you know that the odd ball is one of those eight (obvious). What wasn't as obvious (to me) at first was that you also know, that if the odd coin is later determined to be from the pile which is now on the low side of the scale, IT IS LIGHTER than the rest. Vice versa for the high side.

For whatever reason having to mix the "lighter than", "heavier than", and "known normal" balls was a no brainer once i realized this. Otherwise my solution was the same as Bonanova's noted above.

I really like the array solution though, brilliant Stephcorbin thanks!

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I miss the "what get's wetter as it dries" riddles, i don't even attempt these ones. But still, nice job of accmplishing i didn't even bother to :lol: !

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DEVIDE INTO TWO EQUAL GROUPS......6 & 6........REPEAT STEP FOR LIGHTER SIDE....3&3........THEN TAKE THE SIDE THAT IS LIGHTER AND TAKE 2 ON ONE SIDE AND ADD1 TO THE 1 U HAVE REMAINING .AND DEVIDE THE SIDE THAT IS LIGHTER TO FIGURE OUT WHICH BALL

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The key to the solution is "the least " amount of times you must weigh to determine which ball is heavier or lighter. So it is possible on the first comparison of any two balls you find imbalance. That is the first weighing. By removing one of the two and replacing it with another, by process of elimination you discover the odd ball. If you have equilibrium, it is the first ball you removed. If you have the same imbalance, it is the remaining ball that was not first removed. It is not possible to arrive at the answer with less than two weighings. :)

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