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Altamira_64
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 8 apples   « on: Feb 21st, 2016, 11:16am » Quote Modify

We are given 8 apples of which 2 are rotten; one is lighter than the good apples and the second is heavier, while all the good apples weight the same.
Using a two-arm balance scale and in only 3 weightings, can you tell if the two rotten apples together weight more or equal or less than two good ones? Obviously there are no visible signs of the rotten apples.
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Hippo
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 Re: 8 apples   « Reply #1 on: Mar 9th, 2016, 12:15am » Quote Modify

... My original bet was wrong ...

Suppose apple weights are 100,100,100,100,100,100 and either
A) {98,101},
B) {99,101}, or
C) {99,102}.
Our goal is to detect which of A), B), C) cases  happened.

Let our scalings are as follows:

abcd?efgh
abef?cdgh
abgh?cdef

Let us discuss possible outcomes:

If either of ? is = that mean the weight on both sides is 400 and case B) happened.

Otherwise once the pair of distinct apples were on the same side of the scale and as there was no = it must be either A) or C) case.

Count for each apple pair ab, cd, ef, gh how many times it was on < side and how many times on > side.
In the A) case there is pair (containing 98 ) three times on < side.
In the C) case there is pair (containing 102) three times on > side.

The chosen weights were just for easier explanation.

Edit: I forget 8 could not stand before )
 « Last Edit: Mar 11th, 2016, 9:26am by Hippo » IP Logged
Altamira_64
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Posts: 116
 Re: 8 apples   « Reply #2 on: Mar 9th, 2016, 10:01am » Quote Modify

Yes but how do we know which pair contains 98 (or 102)?
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Hippo
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 Re: 8 apples   « Reply #3 on: Mar 11th, 2016, 9:25am » Quote Modify

on Mar 9th, 2016, 10:01am, Altamira_64 wrote:
 Yes but how do we know which pair contains 98 (or 102)?

Luckilly 98 resp. 102 could be detected (the extreme pair contains the apple), but the task was just to find if its 199,200 or 201 case.
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Altamira_64
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Posts: 116
 Re: 8 apples   « Reply #4 on: Mar 11th, 2016, 11:44am » Quote Modify

From the above 3 cases:

abcd-efgh
abef-cdgh
abgh-cdef
you can have
1. Both rotten apples at the left scale
2. both at the right
3. one plus one

Only by identifying which side is > or <, how can you tell which of the above 3 cases we have? Also, by this, how can you tell if the two bad apples together are lighter or heavier or equal to 2 good ones?
Can you elaborate on this with a diagram?

Many thanks!!
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Hippo
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 Re: 8 apples   « Reply #5 on: Mar 11th, 2016, 1:43pm » Quote Modify

Write inequalities how mesures ended and I will tell you which case it was.

<<< A (3,0) (1,2) (1,2) (1,2) 98\in ab
<<> C (2,1) (2,1) (2,1) (0,3) 102\in gh
<>< C (2,1) (2,1) (0,3) (2,1) 102\in ef
<>> A (1,2) (3,0) (1,2) (1,2) 98\in cd
><< C (2,1) (0,3) (2,1) (2,1) 102\in cd
><> A (1,2) (1,2) (3,0) (1,2) 98\in ef
>>< A (1,2) (1,2) (1,2) (3,0) 98\in gh
>>> C (0,3) (2,1) (2,1) (2,1) 102\in ab
 « Last Edit: Mar 11th, 2016, 1:56pm by Hippo » IP Logged
Altamira_64
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Posts: 116
 Re: 8 apples   « Reply #6 on: Mar 11th, 2016, 2:09pm » Quote Modify

What do you mean by (3,0), (1,2) etc??
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Hippo
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 Re: 8 apples   « Reply #7 on: Mar 11th, 2016, 2:12pm » Quote Modify

on Mar 11th, 2016, 2:09pm, Altamira_64 wrote:
 What do you mean by (3,0), (1,2) etc??

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Altamira_64
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Posts: 116
 Re: 8 apples   « Reply #8 on: Mar 13th, 2016, 1:26am » Quote Modify

You are brilliant!!!

on Mar 11th, 2016, 2:12pm, Hippo wrote: