Question number eight, six superheroes capture 20 villains. The first superhero captures one villain. The second captures two villains and the third captures three villains. The fourth superhero captures more villains than any of the other five. What is the smallest number of villains the fourth superhero must have captured? Let's make a tally of the captured villains as follows. We have 20 of them and they were captured in groups of one, two, and three by the first three superheroes and there is no overlap. Each were a unique capture. The fourth superhero we don't know about. Also the fifth and the sixth have not really reported. So we know that between the three unknown numbers of captured villains there are 14 captures and let's say that the fourth superhero is called X and he must have captured at least four villains and also more than the other two unknown superheroes. And so then we have to go through cases. If X is actually equal to four then Y plus Z amounts to 10 but in that case we can have five plus five as the breakdown and so at least one of Y and Z is bigger than four. So that's not possible. If X captures five villains then Y plus Z is equal to nine and that's equal to either eight plus one which is bad, seven plus two which is bad, six plus three which is bad, five plus four which is bad and then we cycle through those possibilities again. So there is at least a tie at each partition into two. And finally with X equal to six we see that between the remaining two superheroes they have captured eight villains and so eight we can partition into two as six plus two that doesn't work, seven plus one that doesn't work, five plus three that also doesn't work but then finally we have a four plus four which does work. And so supposing that X is six, Y is equal to Z and both of those captured four then we have the smallest number of villains that the fourth superhero must have captured set at six.

superheroes

villains

capture

problem-solving

mathematics