This is in continuation of A Cup Of Coffee.

You have a five cup mug, a three cup mug, a water supply, a sink with a drain, and a packet of instant coffee which when dissolved in one cup of water produces coffee of strength 100%.

The packet may be used at any time, but the entire contents of the packet must be dissolved into a single mug when it is used.

What integer values of c (from 1 to 25 inclusively) is possible if the task is to fix 4 cups of coffee at exactly c% strength? Prove that these are indeed the only possible values of c.

I can find: 25%, 20%, 16%, 8.33%, 6.66%, 2.083%, 1.66%, 1.33% before the mixtures fall below 1% strength. There are two starting possibilities, one where the 3 mug is filled first, and one where the 5 mug is. The 5 mug variant produces 25%, 20% or (with added water to the 20%, and, assuming we are allowed to dip the cup into the final brew) 16%; other possibilities are excluded because the starting 5 is either thrown away or ends up in the 3 mug unable to provide the necessary 4 cuppas.

The 3 mug variant produces 25%, and 8.33% depending when the coffee is added, and additionally 6.67% with added water to the latter in the 5 cup. 

We can have a second shot, starting with any of the brews already made, but because of the restriction already mentioned, the second run has to be with the brew starting in the 3 mug at turn 1, so that only 8.33% of the initial brew emerges at the other end.

The 1% restriction limits the process to 2 shots.

Addendum: Apologies, there are two more possibilities. If we take the 25% brew, we can pour into 3 (discarding 1), pour back into 5, add water, pour back into 3 (discarding 2) for 25%*3/5*8.33%= 1.25%; and the same for 20%*3/5*8.33%= 1% and we need not consider the lesser brews.

So the integer values are: 25%, 20%, (16%), 1%. Then again, maybe not, see subsequent posts!

Edited on March 7, 2012, 1:48 am

Posted by broll on 2012-03-06 13:56:56