ankur rathi

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Ryan McGuire

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Posts: 1113

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Ryan McGuire

Ranch Hand

Posts: 1113

7

posted 10 years ago

For those who think this is too easy, let me add a twist to the problem.

Restriction: You have to know what weighings you will do before you start the whole experiment. In other words, you can not use the results of the first weighing to decide which donuts/marbles to weigh in the second or third one.

Here I'll start you off:

Step one:

Weigh donuts... I mean marbles A, B, C and D against E, F, G and H.

I went through quite a few donuts trying to figure this one out.

Restriction: You have to know what weighings you will do before you start the whole experiment. In other words, you can not use the results of the first weighing to decide which donuts/marbles to weigh in the second or third one.

Here I'll start you off:

Step one:

Weigh donuts... I mean marbles A, B, C and D against E, F, G and H.

I went through quite a few donuts trying to figure this one out.

Jim Yingst

Wanderer

Sheriff

Sheriff

Posts: 18671

posted 10 years ago

I first heard this with coins, back in college. It's not that the original puzzle is too easy, but it's pretty familiar here. Ryan's revision does offer a new challenge. One solution:

1. Weigh abcd (left) vs. efgh (right)

2. Weigh aeij (left) vs. bcfk (right)

3. Weigh chil (left) vs. abgj (right)

Write > if the left is heavier than the right, < if the right is heavier than the left, and = if they balance. Concatenate these three symbols and look up the result below. A + means the given coin/donut/marble is heavier, and a - means it's lighter. An x means the result is impossible.

<<<: x

<<=: f+

<<>: a-

<=<: g+

<==: d-

<=>: h+

<><: c-

<>=: e+

<>>: b-

=<<: i-

=<=: k+

=<>: j-

==<: l-

===: x

==>: l+

=><: j+

=>=: k-

=>>: i+

><<: b+

><=: e-

><>: c+

>=<: h-

>==: d+

>=>: g-

>><: a+

>>=: f-

>>>: x

So if you get 1. left is heavier, 2. right is heavier, 3. they balance, that's ><=, which maps to e-, which means coin/donut/marble e is lighter.

1. Weigh abcd (left) vs. efgh (right)

2. Weigh aeij (left) vs. bcfk (right)

3. Weigh chil (left) vs. abgj (right)

Write > if the left is heavier than the right, < if the right is heavier than the left, and = if they balance. Concatenate these three symbols and look up the result below. A + means the given coin/donut/marble is heavier, and a - means it's lighter. An x means the result is impossible.

<<<: x

<<=: f+

<<>: a-

<=<: g+

<==: d-

<=>: h+

<><: c-

<>=: e+

<>>: b-

=<<: i-

=<=: k+

=<>: j-

==<: l-

===: x

==>: l+

=><: j+

=>=: k-

=>>: i+

><<: b+

><=: e-

><>: c+

>=<: h-

>==: d+

>=>: g-

>><: a+

>>=: f-

>>>: x

So if you get 1. left is heavier, 2. right is heavier, 3. they balance, that's ><=, which maps to e-, which means coin/donut/marble e is lighter.

"I'm not back." - Bill Harding, *Twister*

pradeep jaladi

Ranch Hand

Posts: 65

posted 10 years ago

Hi,

Divide 12 marbles in 6-6. put 6 in one side and remaing six in another let us say if one marble is lesser weigth pick the 6 marbles from the less weigth side and divide six in to 3-3 and weigth again continue the process until 1-1-1 is left. put any 2 out of 3 marbles if they are same reamining marble is weigth less, if any of them is weigth less that will be the weigth less one.

Ex:-

1) 6-6 ( rule out the weigth one)

2) 3-3 ( rule out the weigth one)

3) 1-1-1 ( rule out the original one.)

Divide 12 marbles in 6-6. put 6 in one side and remaing six in another let us say if one marble is lesser weigth pick the 6 marbles from the less weigth side and divide six in to 3-3 and weigth again continue the process until 1-1-1 is left. put any 2 out of 3 marbles if they are same reamining marble is weigth less, if any of them is weigth less that will be the weigth less one.

Ex:-

1) 6-6 ( rule out the weigth one)

2) 3-3 ( rule out the weigth one)

3) 1-1-1 ( rule out the original one.)

Pradeep.Jaladi

Ryan McGuire

Ranch Hand

Posts: 1113

7

ankur rathi

Ranch Hand

Posts: 3830

posted 10 years ago

That�s a kind of universal trick for this kind of problem. But this trick will not work for this problem until we know that the odd marble is either heaver or lighter than others.

Originally posted by pradeep jaladi:

Hi,

Divide 12 marbles in 6-6. put 6 in one side and remaing six in another let us say if one marble is lesser weigth pick the 6 marbles from the less weigth side and divide six in to 3-3 and weigth again continue the process until 1-1-1 is left. put any 2 out of 3 marbles if they are same reamining marble is weigth less, if any of them is weigth less that will be the weigth less one.

Ex:-

1) 6-6 ( rule out the weigth one)

2) 3-3 ( rule out the weigth one)

3) 1-1-1 ( rule out the original one.)

That�s a kind of universal trick for this kind of problem. But this trick will not work for this problem until we know that the odd marble is either heaver or lighter than others.

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