Riddle about 3 men competing and never tying

Question

Mr. A, B, C decided compete in athletics against each other. They competed in few disciplines.

Winner got x points, second y points, last z points: x>y>z. x,y,z are integers. There were no draws(ties) in any discipline.

Mr. A scored 22 points, Mr. B and Mr. C scored 9 points. Mr. B won the long jump. Who was the second in the 1/4 mile run?

I made a model:

A - 22 points B - 9 points C - 9 points

B won the long jump

$$ MAXPOINTSTOSCORE = 22+9+9=40 $$

I can't guess 3 integers(x,y,z) which would match this data.

Answer 1

The answer is Mr.C. In detail following things happened:

1. They competed in $5$ disciplines.
2. The values of $x,y,z$ are $(5,2,1)$ respectively.
3. Mr. A came first in $4$ disciplines and second in long jump, thus collecting $4\times5+2=22$ points. Note that this means that Mr. A won the 1/4 mile run.
4. Mr. B came first in the long jump and then (necessarily) finished third in all other disciplines, thus collecting $5+4\times1=9$ points. Note that this means that Mr. B finished third in the 1/4 mile run.
5. Mr. C came third in the long jump and (necessarily) second in every other discipline. thus collectiong also $1+4\times2=9$ points.

The justification is presented below.

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Denote with $N$ the number of disciplines in which they competed. Of course $N$ is an integer. Then you know that $$N(x+y+z)=40=5\cdot2^3$$ which means that $$N\in\{5,10,20, 2,4,8\}$$ and similarly $$x+y+z \in \{5,10,20,2,4,8\} $$But since $x>y>z>0$ then you have that $x+y+z\ge3+2+1=6$ so that $$x+y+z\in \{8,10,20\}$$ and therefore $$N \in \{5, 4, 2\}$$ Moreover, since Mr. B has won at least one discipline and has $9$ points then we have that $x\le 8$ because they competed in at least two disciplines (see values of $N$). But then it is impossible for Mr. A to have been able to collect $22$ points if there are only two disciplines! So, the possible values of $x+y+z$ and $N$ are further reduced to $$x+y+z\in\{8,10\}$$ and $$N\in\{5,4\}$$ respectively. Now, since there are only two cases, we can treat them separately

1. Case: $N=5, x+y+z=8$. Then necessarily $x=5,y=2,z=1$.

2. Case: $N=4, x+y+z=10$. Then $x=7,y=2,z=1$ or $x=6, y=3, z=1$ or $x=5, y=4, z=1$ or $x=5, y=3, z=2$. Now, we take all these triplets to see if we can create the numbers $22$ and $9$ with combinations of $4$ of the given values.

   • If $x=7$ then necessarily Mr.A won $3$ disciplines and came 3rd in the remaining 4th discipline. But, then Mr. B won the 4rth discipline (long jump) so that Mr. C won no discipline. But then the best Mr.C can do is to collect $4\cdot y=4\cdot 2=8<9$ points. So this case is excluded.
   • If $x=6$ then you cannot reach $22$ with any combination of $x,y,z$ (using $4$ of them). So this case is also excluded.
   • If $x=5$ similarly. So the last two triplets of the case $N=4$ are excluded!

In sum, we found that the only admissible values of $x,y,z,N$ are $$x=5, y=2, z=1$$ and $N=5$. Now, it is obvious that Mr. A won 4 out the 5 disciplines and came second in the remaining one, which was long jump and which was won by Mr. B. But, then Mr.B collected in total $9$ points by finishing third in every other discipline, because in any other case he would have collected more than $9$!! So Mr.C is the one who came second in every other discipline except for the long jump where he came third.

Answer 2

I think it's C. Let the points be 5 for first, 2 for second, and 1 for third, with 5 events total. Then A won all events except the long jump, where he got second. B was last in all events except the long jump, which we know he won. And C was second in every event except the long jump, where he was last. So if C was second in everything else, he must have been second in the 1/4 mile run.