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Guest Message by DevFuse

# Weighing machines

Best Answer DeGe, 26 August 2014 - 11:36 AM

Come to think of it...

Spoiler for Old is gold
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14 replies to this topic

### #11 DeGe

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Posted 04 September 2014 - 08:54 AM

If the caretaker remembers 5 eggs which he might have swapped two of them,
(5c2) or 10 possible combinations,how he could find the pair with 2 trials?

The solutions above seem to be for the first 5 eggs. I understood that it was any 5 eggs out of the 12. Try this then!

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### #12 TimeSpaceLightForce

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Posted 04 September 2014 - 05:57 PM

Spoiler for

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### #13 k-man

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Posted 18 September 2014 - 09:25 PM

I'm not going to spoiler this as I'm not providing any specific solutions. I've been fiddling around with this problem for some time and still couldn't find a generic formula for digital scale that would determine a number of weighings necessary for a given number of eggs.

However, I can definitely solve the original problem with 12 eggs in 4 weighings using a digital scale.

If limited to 3 weighings, then only up to 9 eggs (36 pairwise combinations) can be solved.

10 eggs (45 pairwise combinations) requires 4 weighings, but if we can reduce the number of combinations by 1 (down to 44) then it can be solved in 3 weighings.

So, if anyone is interested in solving this variation then try this:

Among 10 eggs two are switched, but you know that it's definitely not 1 and 2 that are switched. Find the switched pair using only 3 weighings with a digital scale.

Edited by k-man, 18 September 2014 - 09:28 PM.

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### #14 TimeSpaceLightForce

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Posted 19 September 2014 - 12:31 PM

that's the squeeze i was looking for..nice

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### #15 TimeSpaceLightForce

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Posted 26 September 2014 - 09:12 PM

Among 10 eggs two are switched, but you know that it's definitely not 1 and 2 that are switched. Find the switched pair using only 3 weighings with a digital scale.

Spoiler for

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