1 problem found
The Tour de Clochemerle is not yet as big as the rival Tour de France. This year there were five riders, Arouet, Barthes, Camus, Diderot and Eluard, who took part in five stages. The winner of each stage got 5 points, the runner up 4 points and so on down to the last rider who got 1 point. The total number of points acquired over the five states was the rider's score. Each rider obtained a different score overall and the riders finished the whole tour in alphabetical order with Arouet gaining a magnificent 24 points. Camus showed consistency by gaining the same position in four of the five stages and Eluard's rather dismal performance was relieved by a third place in the fourth stage and first place in the final stage. Explain why Eluard must have received 11 points in all and find the scores obtained by Barthes, Camus and Diderot. Where did Barthes come in the final stage?
Solution: Since \(A\) scored \(24\) points, he must have finished first in all but one race and second in that race. Given \(E\) won the final stage, \(A\) must have been \(11112\) \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & \\ C & - & - & - & - & - & \\ D & - & - & - & - & - & \\ E & - & - & - & 3 & 1 \\ \end{array} If \(E\) has \(12\) points the smallest number of points the others can have are \(13, 14, 15\) which would be a total of \(78\) points \(\geq 15 \times 5 = 75\), more than is available, therefore \(E\) must have the minimum \(11\) points. \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & \\ C & - & - & - & - & - & \\ D & - & - & - & - & - & \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array} There are now \(40\) points to be divided between \(B, C\) and \(D\). \(12+13+14 = 39\), so only way to achieve this is \(12, 13, 15\). \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & 15 \\ C & - & - & - & - & - & 13 \\ D & - & - & - & - & - & 12 \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array} Camus gained the same position in four of the five races. So we need \(4x + y = 13\) which can be done with \(4 \times 1 + 9\) or \(4 \times 2 + 5\) or \(4 \times 3 + 1\). The first two aren't possible (you can't score \(9\)) and the second isn't possible (all the first places are taken) so \(C\) must have four third places and a last place. (Which also must be the second to last race since there are alread last places in \(3\) of the races and a third place in the second to last) \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & 15 \\ C & 3 & 3 & 3 & 5 & 3 & 13 \\ D & - & - & - & - & - & 12 \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array} There are now one \(5\), five \(4\)s, four \(2\)s left to place. And they need to add to \(12\) for one rider. [In score terms this is \(1, 5\times 2, 4 \times 4\). Neither rider can have all the second places, and since they would score too highly, and \(D\) can't have more than one second place since otherwise he'd score too highly. Therefore \(B\) has three second places. So \(B\) is \(1,2,4,4,4\) and \(C\) is \(4,2,2,2,2\) in some order. \(D\) can't come second in the last race, so he comes \(4\)th and \(B\) comes \(5\)th \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & 5 & 15 \\ C & 3 & 3 & 3 & 5 & 3 & 13 \\ D & - & - & - & - & 4 & 12 \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array}