Daily logic 5

At first sight, this seems completely impossible; we have many unknowns, and hardly any information.

However, the crucial information we do have is that we are dealing in lots of cases with small, positive integers, and, surprisingly, that's enough.

Let's start by figuring out something about $n$ , the number of events there were. In every event, a total of $x+y+z$ points are awarded. So overall, $n(x+y+z)$ points are awarded. But we know that overall, $22+9+9=40$ points are awarded; so $n(x+y+z)=40$ and $n$ is a factor of $40$ .

Since $z>0$ , every player scores at least one point in every event. So there cannot have been more than $9$ events, since both Bob and Charlie score $9$ points each.

In fact, Bob wins an event, so his score must be at least $x+(n-1)z \ge 3+n-1 = n+2$ ( $x$ is at least $3$ , $z$ is at least $1$ ). So there must have been fewer than $7$ events.

Two events (100m, javelin) are mentioned in the problem; so $n\ge2$ .

Let's say these were the only two events. Adam wins one, Bob the other. Since Adam's final score is higher than Bob's, Adam must have come second in the javelin (Bob's winning event) and Bob must have come third in the 100m (Adam's winning event). This would mean that:

$A=22=x+y$ , $B=9=z+x$ , $C=9=y+z$ . But the last two equations imply that $x=y$ , which is not the case.

So we've narrowed the number of events down to either four or five.

Let's say there were four events.

The most Adam could score is by winning three events and coming second in the javelin. So we get $22\le 3x+y < 4x$ ; hence $x>5$ . The least Bob could score is one win in the javelin and three last places - so $9 \ge x+3z \ge x+3$ ; hence $x\le 6$ , and combined with what we worked out from Adam's score, we must have $x=6$ and $z=1$ . For Adam to score $22$ , he must in fact come second in the javelin and first in everything else, and we must have $y=22-3\times6=4$ .

Under the assumption that there are four events, we've found that $(x,y,z)=(6,4,1)$ , and that all events are won by either Adam or Bob. But how can Charlie score $9$ points? It's not possible, so we reach a contradiction, and we have worked out that there must have in fact been five events. (Another contradiction is that $x+y+z=11$ is not a factor of $40$ ).

We can use the same steps as above to find out the points awarded. Once again, Adam's score is maximised if he wins every event except the javelin, coming second in that. So $22 \le 4x+y < 5x$ ; hence $x>4$ . Bob's score is minimised (again) if he wins the javelin and places third in everything else - so $9 \ge x+4z \ge x+4$ and $x \le 5$ . Once again, only one value works, and we have $x=5$ and $z=1$ (and it is the case that Bob is last in every event except javelin). For Adam to score $22$ , we must have $y=2$ .

This time the scores work out: the table of points is below

As we've shown, this set of results is unique; and Charlie must have come second in the 100m.