
Originally Posted by Monique

Ok, let's say Barry the Baboon has a tetrahedron with n balls on every edge. Three sides of this tetrahedron are visible, it means he has to clean only the balls which are on these three sides.
On the first side there are n(n+1)/2 dirty balls.
On the second side, the balls on one edge are already cleaned, so there are n(n+1)/2 -n dirty balls.
On the third side the balls on two edges are cleaned, so there are n(n+1)/2 -2n+1 dirty balls.
If we add up all the dirty balls, we see that in total Barry the Baboon has to clean 3n(n-1)/2 +1 balls.
Now, let y be the number of yellow balls on an edge and let b be the number of brown balls on an egde. Then
3y(y-1)/2 +1 - 3b(b-1)/2 +1 = 45.
The number of brown balls on the edge was 1 less than the number of yellow balls on the edge, so we can replace b by y-1 and after some calculations we get that y=16, i.e.
there were 16 balls on an edge of the tetrahedron of yellow balls and on the three visible sides there were 361 balls.
As it took 1 minute to clean one ball, Barry the Baboon had to waste 361 minutes or 6 hours and 1 minute to clean all the dirty yellow balls. BTW, it will take 5 hours and 16 minutes to clean the dirty brown balls... of course he was angry!
Sorry for my English!

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