The odd stone logic problem

Alan McDougall · #1 02-02-2009, 09:15 AM

Hi,

Especially for XRIS who can reconcile every problem by logic I hope this problem had not already being tried by the forum ,if not it is an excellent test of deduction

Can you solve this problem?

There are 12 balls of exact volume and color 11 of equal weight/mass, 1 of minutely different unequal weight / mass(either heavier or lighter).

The difference is too small to feel by hand

Use 3 weighings on a balance scale to tell what stone has a different weight and if it is heavier or lighter.??

Scale libra type balance scale you are not allowed to use a bathroom type scale

xris · #2 02-02-2009, 10:49 AM

Quote:

Originally Posted by Alan McDougall

Hi,

Especially for XRIS who can reconcile every problem by logic I hope this problem had not already being tried by the forum ,if not it is an excellent test of deduction

Can you solve this problem?

There are 12 balls of exact volume and color 11 of equal weight/mass, 1 of minutely different unequal weight / mass(either heavier or lighter).

The difference is too small to feel by hand

Use 3 weighings on a balance scale to tell what stone has a different weight and if it is heavier or lighter.??

Scale libra type balance scale you are not allowed to use a bathroom type scale

i can do it easy with four have to think about it.

xris · #3 02-02-2009, 11:03 AM

I think i have it four each side..one side is heavier..change on the lighter side two of the balls for two of the balls not used..if still out of balance exchange one of the balls of the two balls left on the the side you have just changed with one on the other side. If it was anything else do virtualy the same thing.ie.four and four balanced..exchange two balls then it would be out of balance..exchange one for one of the four balls you have just put there.

Alan McDougall · #4 02-02-2009, 01:26 PM

XRIS,

Quote:

I think i have it four each side..one side is heavier..change on the lighter side two of the balls for two of the balls not used..if still out of balance exchange one of the balls of the two balls left on the the side you have just changed with one on the other side. If it was anything else do virtualy the same thing.ie.four and four balanced..exchange two balls then it would be out of balance..exchange one for one of the four balls you have just put there.

But XRIS if you compare four balls like you stated and they do not balance you don't know at this stage whether the upper pan contains the lighter odd ball or if the odd ball is heavier and in the lower pan

They odd ball might be heavier of it might be lighter , you don't know where it is

The solution must be fool proof (no pun intended)

xris · #5 02-02-2009, 03:06 PM

Quote:

Originally Posted by Alan McDougall

XRIS,

But XRIS if you compare four balls like you stated and they do not balance you don't know at this stage whether the upper pan contains the lighter odd ball or if the odd ball is heavier and in the lower pan

They odd ball might be heavier of it might be lighter , you don't know where it is

The solution must be fool proof (no pun intended)

Listen ..i have four balls one side and four the other now you decide if they are balanced or not balanced...the other four balls are waiting for your decision..

MJA · #6 02-02-2009, 03:47 PM

Solution!

If they have the same volumn then the one different ball can only be heavier.
Thusly

Measurement 1: Divide balls 6 X 6 and weigh.

Measurement 2: Divide Groupe of 6 heaviest balls into 3 X 3 and weigh.

Measurement 3: Weigh 2 of the remaning balls of the heavy group of three, if even the third ball is the heaviest. If uneven the solution is also known.

That took some valuable time to figure out,
Thanks!!!

=
MJA

xris · #7 02-02-2009, 04:03 PM

Quote:

Originally Posted by MJA

Solution!

If they have the same volumn then the one different ball can only be heavier.
Thusly

Measurement 1: Divide balls 6 X 6 and weigh.

Measurement 2: Divide Groupe of 6 heaviest balls into 3 X 3 and weigh.

Measurement 3: Weigh 2 of the remaning balls of the heavy group of three, if even the third ball is the heaviest. If uneven the solution is also known.

That took some valuable time to figure out,
Thanks!!!

=
MJA

But one could be heavier or lighter..The other group could be the one thats lightest.

Zetherin · #8 02-02-2009, 10:31 PM

Quote:

Originally Posted by MJA

Solution!

If they have the same volumn then the one different ball can only be heavier.
Thusly

Measurement 1: Divide balls 6 X 6 and weigh.

Measurement 2: Divide Groupe of 6 heaviest balls into 3 X 3 and weigh.

Measurement 3: Weigh 2 of the remaning balls of the heavy group of three, if even the third ball is the heaviest. If uneven the solution is also known.

That took some valuable time to figure out,
Thanks!!!

=
MJA

Are you sure that only counts as 3 weighings? That seems like 6 to me. You'd have to take one of the 6's off and then weigh the other 6 (meaning two seperate weighings). Likewise, with the 3's and then the remaining 2. =6 weighings.

MJA · #9 02-02-2009, 10:55 PM

1: Divide the circles evenly and weigh them.

OOOOOO OOOOOO

V

2. Divide the heaviest side again and weigh again.

OOO OOO

V

3. Weigh two of the circles, if even the third circle is the solution.

If not, the solution is again known.

O O

V

O

=

MJA

Zetherin · #10 02-02-2009, 11:18 PM

Quote:

Originally Posted by MJA

1: Divide the circles evenly and weigh them.
OOOOOO OOOOOO
V
2. Divide the heaviest side again and weigh again.
OOO OOO
V
3. Weigh two of the circles, if even the third circle is the solution.
If not, the solution is again known.
O O
V
O
=
MJA

My fault, I didn't see that it was a balance scale, implying you can have two weighing at the same time.

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The following users say: THANK YOU - Alan McDougall for the above post!
MJA