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Cannonpointer
26 Aug 2022 4:12 pm
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98% Macho Man
98% Macho Man
47,166 posts
Okay, I'm back. I have solved for the initial 8 being control, and I have solved for the sequestered 4 being control, but only when sequestering 2 from a and 1 from b and getting an even weight. An odd weight defeated me. 

An odd weight left us with 5 suspects in 2 groups of 4 - a and b, light and heavy, respectively. Two are in one group, and three are in the other. We have identified neither the suspect nor the weight problem, and we have only one weighing left. 

I need an applied math course - I should be able to set this up. I don't now whether I need to back out of my second solution (Replacing 2&1 of the suspects from the control group of four), or think harder about the present problem in the event of an unequal weighing (Leaving us with 3 suspects on one side and 2 on the other - and 9 control marbles identified). Hmmm...

Musing...

I simply cannot see any way to work out the 3v2 dead end in a single weighing... so I would have to go back and try something OTHER than replacing 1 from a and 2 from b...

That is another dead end, for me. I am going to don my softest leather loafers - **** that, my cloth slippers - and prepare to kick my own ***. @RunningWithScissors, @Jantje_Smit, I surrender. Image

 
 
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