Find the position of sprinters in according to their vest numbers $\text{Sprinters in Olympics running on the track are carrying} \\ 
\text{numbers from 1 to 7 at the back of their vests. They} \\ 
\text{complete the race when their feets touches the finish} \\ 
\text{line at 9.10, 9.20, 9.30 seconds till 9.7 seconds i.e. } \\ 
\text{an interval of one mili second each. Each athelete } \\ 
\text{is ranked from 1 (for the fastest) to 7 (for the slowest).} \\ 
\text{The following additional information is known:} \\ 
\underline{\ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\
\text{(i) The square of the highest total of the vest number} \\ 
\text{and rank is 169 which occurs inly once.} \\ 
\text{(ii) The square of the lowest total of vest number} \\ 
\text{and rank is 16 and occurs only once.} \\ 
\text{(iii) The square of the total of vest number and} \\ 
\text{rank = 81, occurs thrice.} \\ 
\text{(iv) The winner's vest no. exceeds that of first
runner up.}\\
\bf \text{Draw the correct table with rank according to their vest numbers.}$
From the in formation given I have drawn the following conclusions.
From statement $\ (i)$
$\begin{array}{|c|c|c|} \hline
\text{Rank} & \text{Vest no.} \\ \hline
6 &  7  \\ \hline
7 &6    \\ \hline
\end{array}$
From statement $\ (iii)$
$\begin{array}{|c|c|c|} \hline
\text{Rank} & \text{Vest no.} \\ \hline
2 & 7   \\ \hline
3 & 6   \\ \hline
4 & 5   \\ \hline
5 & 4   \\ \hline
6 & 3   \\ \hline
7 & 2   \\ \hline
\end{array}$
From statement $\ (ii)$
$\begin{array}{|c|c|c|} \hline
\text{Rank} & \text{Vest no.} \\ \hline
1 &  3  \\ \hline
2 &2    \\ \hline
3 & 1   \\ \hline
\end{array}$
I am confused on how to proceed further.
I look for a short and a simple way.
I have studied maths upto $12$th grade.
 A: There is no simple solution. You have to argue by cases and eliminate all the possibilities that are wrong, leaving only one possible solution.
For conciseness use the notation $r1, r2,\ldots, r7$ to represent the ranks of vests 1 through 7. So for example, $r3=1$ means that vest 3 was the winner.
The solution arises from considering the case $r1=3$ and $r7=6$. If this assignment is correct, then this rules out possibilities $r7=2$, $r6=3$, and $r3=6$ from (iii), and therefore the remaining three statements $r5=4$, $r4=5$, $r2=7$ must be true. It remains to assign ranks 1 and 2 to the remaining vests 6 and 3. Since (ii) rules out $r3=1$, we must have $r3=2$ and finally $r6=1$.
You have to eliminate all the other possibilities:


*

*$r3=1$

*$r2=2$

*$r1=3$ and $r6=7$


For example, we can eliminate the first possibility by arguing: Suppose $r3=1$. Then by (iv) either $r2=2$ or $r1=2$. By (ii), we must have $r1=2$. This now rules out $r7=2$ and $r3=6$ from (iii), leaving four possibilities: $r6=3, r5=4, r4=5, r2=7$, exactly three of which are true. If $r6=3$ is the false one, then $r6=6$ is the only other choice for $r6$, contradicting (i). Therefore $r6=3$ is true, which means exactly two of the statements $$r5=4, r4=5, r2=7\tag{*}$$ are true. But since $r6=3$, (i) implies $r7=6$. So now the only unassigned ranks are 4, 5, 7; these must be assigned in some order to vests 5,4,2. But there is no arrangement of these ranks that makes exactly two of the statements in (*) true and the other false. It follows that $r3=1$ is impossible.
