Yesterday on Puzzling SE a puzzle was posted, see here
I didn't get the solution posted there so I am rewriting it for discussing it here.
121 women are competing in the heptathlon in Olympic Games.
If British track and field star Jessica Ennis finishes in 104th place in every event, what is the highest place she can finish in the overall heptathlon?
Heptathlon comprises of $7$ events.
Suppose there are no ties for any position in any of the events (each position is occupied by exactly one participant ) and suppose the position $1$ is awarded maximum points ( $ 121 $ ) while the last position is awarded least points ( $ 1 $ ), this way Jessica would have scored $18 \times 7 = 126$ points at the end of seven events, and if we consider that the participants coming after Jessica are all different for each event, there will be $17 \times 7 = 119$ participants whom Jessica beat in total seven events.
With this much information, is it possible using High School Algebra and Logic to arrive at some conclusion for this problem? Can we minimise the total points earned by the other participants using High School Algebra?
I believe this problem can be solved by a persistent high school student, if he has enough time. I don't think it requires any particularly insightful combinatorics.
Jessica's average place is $104$. For her to beat someone, their average place must be greater than $104$.
The highest average place a competitor can make is $121$, then the next $120$, then the next $119$, etc. The optimal situation for Jessica is for all the same group of $P$ players to all share the worst scores. Suppose $m$ is the best rank that anyone who ultimately lost to Jessica got in an individual event. Then:
$$\text{Jessica's average} < \text{Worst Player's Average}$$ $$104 < \frac{1}{P} \sum_{k=m, k \ne 104}^{121} k$$ $$104 < \frac{1}{121 - m} \left( \frac{121(121 + 1)}{2} - \frac{(m-1)m}{2} - 104 \right)$$ $$m > 87$$
So if it is possible to setup players to all get ranks $88,~ 89,~ \dots ,~ 102,~ 103,~ 105,~ 106,~ \dots ,~ 120,~ 121$ and still have an average worse than $104$, then Jessica gets ranked $88^{\text{th}}$.
This can be done with putting the students into groups of 7 (and one group of 5):
Group 1 all got scores of : $\{88,~ 89,~ 90,~ 102,~ 119,~ 120,~ 121\}$, for an average of $104.14$
Group 2 all got scores of : $\{91,~ 92,~ 94,~ 101,~ 116,~ 117,~ 118\}$, for an average of $104.14$
Group 3 all got scores of : $\{93,~ 95,~ 96,~ 103,~ 113,~ 114,~ 115\}$, for an average of $104.14$
Group 4 all got scores of : $\{97,~ 98,~ 99,~ 109,~ 110,~ 111,~ 112\}$, for an average of $105.14$
Group 5 all got scores of : $\{100, 105, 106, 107, 108\}$, setup as:
$$\begin{array} {c|ccccc} & \text{Player 1} & \text{Player 2} & \text{Player 3} & \text{Player 4} & \text{Player 5} \\ \hline \text{Event 1} & 100 & 105 & 106 & 107 & 108 \\ \text{Event 2} & 105 & 100 & 105 & 106 & 107 \\ \text{Event 3} & 106 & 105 & 100 & 105 & 106 \\ \text{Event 4} & 107 & 106 & 105 & 100 & 105 \\ \text{Event 5} & 108 & 107 & 106 & 105 & 100 \\ \text{Event 6} & 100 & 108 & 107 & 106 & 105 \\ \text{Event 7} & 105 & 100 & 108 & 107 & 106 \\ \hline \text{Average:} & 104.43 & 104.43 & 105.29 & 105.14 & 105.29 \end{array}$$
So it is possible for Jessica to get $88^{\text{th}}$ place despite ranking $104^{\text{th}}$ place in every event. I think it would be difficult for a high school student to come up with the rankings that give her that placement, but it is more a matter of persistence than insight.
The above assumes that best ranked means "beat the most number of opponents". But, if you are satisfied by being tied for last place with all the other last place players, then it is possible to tie for $87^{\text{th}}$ place.
Group 1 : $\{87,~ 88,~ 96,~ 97,~ 119,~ 120,~ 121\}$
Group 2 : $\{87,~ 89,~ 90,~ 91,~ 111,~ 117,~ 118\}$
Group 3 : $\{92,~ 93,~ 98,~ 110,~ 114,~ 115,~ 116\}$
Group 4 : $\{94,~ 95,~ 99,~ 108,~ 109,~ 110,~ 113\}$
Group 5 :
$$\begin{array} {c|ccccc} & \text{Player 1} & \text{Player 2} & \text{Player 3} & \text{Player 4} & \text{Player 5} & \text{Player 6} \\ \hline \text{Event 1} & 101 & 102 & 103 & 105 & 106 & 107\\ \text{Event 2} & 102 & 103 & 105 & 106 & 107 & 101\\ \text{Event 3} & 103 & 105 & 106 & 107 & 101 & 102\\ \text{Event 4} & 105 & 106 & 107 & 101 & 102 & 103\\ \text{Event 5} & 106 & 107 & 101 & 102 & 103 & 105\\ \text{Event 6} & 105 & 103 & 101 & 106 & 102 & 107\\ \text{Event 7} & 106 & 102 & 105 & 101 & 107 & 103\\ \hline \text{Average:} & 104 & 104 & 104 & 104 & 104 & 104 \end{array}$$
And you and $7 \times 4 + 6$ people get last place all with average ranks of $104$, a $35$ way tie making your overall rank $87^{\text{th}}$.