A Weighty ball problem?

zoombini · 07-07-2004, 13:13

Okidoki 8 ball bearings, one of which is slightly different from the rest.
You cannot visually see the difference its only in its weight.
You do not know if its heavier or lighter.

You have a pair of balance scales, you are only allowed to make 3 balance tests.

How do you tell which is the odd bearing?

Something else thats being discussed in the office.

Quick replies wanted so I can be the one to answer it

altis · 07-07-2004, 13:19

2^3 = 8 so, yes, it can be done in three tests.

1) ABC vs DEF
2) ABF vs DEC
3) AEF vs DBC

then just follow the letters...

Edit
Just read the bit about not knowing if it's heavier of lighter

mmmm... thinking cap on, instead.

danielf · 07-07-2004, 13:21

Put 2 bearings on each side of the scale. If they are not equal in weight, the odd one is in this lot. Now check two of these, if these are equal in weight, check one of these against one of the other two. If these are equal in weight, the remaining one is the odd one out. If not, it's the 'new one'. If the first 4 are equal in weight, do the last steps with the other 4.

Edit: Or, in letters. Check AB vs CD. If unequal, check A v B, and the B v C.
If AB = CD repeat last steps with E, F ,G and H.

zoombini · 07-07-2004, 13:22

eh?
what about g & h?

zoombini · 07-07-2004, 13:24

Quote:

Originally Posted by danielf

Put 2 bearings on each side of the scale. If they are not equal in weight, the odd one is in this lot. Now check two of these, if these are equal in weight, check one of these against one of the other two. If these are equal in weight, the remaining one is the odd one out. If not, it's the 'new one'. If the first 4 are equal in weight, do the last steps with the other 4.

You only get 3 goes.

danielf · 07-07-2004, 13:27

Quote:

Originally Posted by zoombini

You only get 3 goes.

This is three goes. In the first go, you narrow it down to four (say ABCD). Then you check two of these to see if they are equal (say A and B). If they are, you only have two left, one of which includes the odd one out. If you now check if A and C are equal (and they are), the odd one out must be D. If A and C are not equal, it must be C.

If the first 4 are equal, you do step 2 and 3 with EFGH

MetaWraith · 07-07-2004, 13:28

Quote:

Originally Posted by zoombini

Okidoki 8 ball bearings, one of which is slightly different from the rest.
You cannot visually see the difference its only in its weight.
You do not know if its heavier or lighter.

You have a pair of balance scales, you are only allowed to make 3 balance tests.

How do you tell which is the odd bearing?

Something else thats being discussed in the office.

Quick replies wanted so I can be the one to answer it

This is an OLD,OLD Problem

Big hint here : Start with weighing 2 vs 2

Full answer here
http://www.mathnerds.com/mathnerds/b...t/solution.asp

danielf · 07-07-2004, 13:30

Quote:

Originally Posted by MetaWraith

This is an OLD,OLD Problem

Big hint here : Start with weighing 2 vs 2

Quote:

Originally Posted by danielf

Put 2 bearings on each side of the scale. <snip?

altis · 07-07-2004, 13:38

Too much haste - ignore me I'm talking balls!

I think Danielf has the answer:

1) AB vs CD, if they balance goto 5
2) A vs B, if they balance goto 4
3) A vs C, if they are unbalanced then answer is A, balanced it's B
4) A vs C, if they are unbalanced then answer is C, balanced it's D
5) AB vs EF, if they balance goto 7
6) A vs E, if they are unbalanced then answer is E, balanced it's F
7) A vs G, if they are unbalanced then answer is G, balanced it's H.

Matth · 07-07-2004, 13:40

1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference.
3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one

danielf · 07-07-2004, 13:46

Quote:

Originally Posted by altis

Too much haste - ignore me I'm talking balls!

I think Danielf has the answer:

1) AB vs CD, if they balance goto 5
2) A vs B, if they balance goto 4
3) A vs C, if they are unbalanced then answer is A, balanced it's B
4) A vs C, if they are unbalanced then answer is C, balanced it's D
5) AB vs EF, if they balance goto 7
6) A vs E, if they are unbalanced then answer is E, balanced it's F
7) A vs G, if they are unbalanced then answer is G, balanced it's H.

Sort of. If AB and CD balance, the odd one out must be in EFGH, if not it must be in ABCD. If ABCD contains the odd one out, compare A with B. If they balance, the odd one out must be C or D. If A and C balance, it must be D. If A and C don't balance, it must be C since A balances with B. If A and B don't balance, one of them is the odd one out, so you compare A (or B with C. Three steps.

danielf · 07-07-2004, 13:52

Quote:

Originally Posted by Matth

1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference.
3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one

Slightly different from my solution, but it works just the same.

zoombini · 07-07-2004, 13:53

The links given are for a 12 ball similar problem.

zoombini · 07-07-2004, 14:02

Quote:

Originally Posted by Matth

1: Weigh 2 against 2 - if this group are equal, they become your reference, the other group the sample, otherwise reverse it.
2: Weigh 2 of the sample against 2 of the reference - if equal, they move to the reference, otherwise the other two move to the reference.
3. Now out of the remaing two left in the sample, weigh ONE against one of the reference - if unequal, you've found it, otherwise it's the other one

This is the easiest to understand for me & it seems to work.

I'll have to see what the lads here make of it. cheers.

MetaWraith · 07-07-2004, 14:04

Now I've had time to take lunch, and write it up here's a solution
I think It's more or less what Danielf had, but with more detail.

STEP 1
Weigh 2 v 2
Possible results
A) pairs are unequal (heavier or lighter not known)
B) pairs are equal
Thus we have 4 balls known to be "good" and 4 balls still "suspect".

STEP 2)
Weigh 2 "suspect" balls v 2 "known good" balls
Possible results
C) pairs are unequal (heavier or lighter is now known)
D) pairs are equal (discard these)

STEP 3C
Weigh one of the "suspect" balls tested in STEP 2 against a "known good" ball
Possible Results
E) Balance equal (the other suspect is guilty)
F) Balance unequal (the chosen suspect is guilty)

STEP 3D
Weigh one of the "suspect" balls not tested in STEP 2 against a "known good" ball
G) Balance equal (the other suspect is guilty)
H) Balance unequal (the chosen suspect is guilty)

NOTE
In G) it is impossible to tell if the guilty ball is heaver or lighter without a further weighing