Re: Friday Puzzler --Track Meet
Larry Barrett
You do not say how many events there are, so that is another variable.
As an example, if there are just three events, and if B wins the javelin and comes in third in the other two, then the most x can be is 7 and the least z can be is 1. B wins the javelin and places third in the other two for 7+1+1=9 points. If C places third in the javelin for 1 point, and comes in second in the other two then in order for C to have 9 points, 2y must equal 8, or y=4, so C will have a total of 1+4+4=9 points. A wins the other two events and places second in the javelin for 7+7+4=18 points. Not enough points for A. And by observation, any other set of values for x, y, z would have fewer points for x and would result in a lower score for A.
So there must be more than three events. With 4 events, the most x can be is 6 and the least z can be is 1. B would have to come in third in the other three events for a total score of 6+1+1+1=9. C could come in second in three events and third in the javelin. But then 3y would have to equal 8; y would not be an integer so this set of values does not work. And any lower value for x would mean that A could not reach 22 points.
With 5 events, the most x can be is 5 and the least z can be is 1. B wins the javelin and comes in third in the other 4 events for 5+1+1+1+1=9 points. A wins the other 4 events for a total of 4x5=20 points. If A comes in second in the javeiin and y=2, then A has 22 points. That means C comes in second in the other 4 events, including the 100 yd dash, and third in the javelin for a total of 2+2+2+2+1=9 points, which satisfies all the point totals.
I think this is one solution. There may be others.
I vaguely remember from an algebra class that equations for which the only solutions are integers are called diophantine equations. This may be an example of a problem where a diophantine equation could be derived involving x,y,z and n. Or, Alex will come up with a clever way of looking at this that makes the solution obvious.
