5. Magnus and Viswanathan play a game against each other three times:
- In each game, each player's score is a positive integer. The players' scores at the end of the three games are six different integers.
- In each game, the winner's score is at least 25 points. If the winner's score is 25 , their opponent's score must be at most 23 . If the winner's score is greater than 25 , their opponent's score must be exactly 2 less than the winner's score.
- Viswanathan wins either the first game or the second game, but not both.
- Viswanathan wins the third game with a score of 25 .
- Each player's score in the second game is the average of their scores in the first and third games.
What was Magnus's score in the third game?
5. When three numbers $a, b, c$ have the property that the middle number, $b$, is the average of the other two, then $b=\frac{a+c}{2}$ tells us that $2 b=a+c$ or $b+b=a+c$ and so $b-c=a-b$.
In other words, if $b-c=d$ (that is, if $b=c+d$ for some number $d$ ), we have $a-b=d$, which gives $a=b+d=(c+d)+d=c+2 d$.
We note that the common difference $d$ could be positive or negative.
Since Viswanathan's three scores have this property and his score in the third game is 25 , then his scores in the first two games can be written as $25+2 d$ and $25+d$, as shown in the table:
\begin{tabular}{c|c|c|c|} 
& 1st game & 2nd game & 3rd game \\
\hline Viswanathan & $25+2 d$ & $25+d$ & 25 \\
\hline Magnus & & & \\
\hline
\end{tabular}

Since Viswanathan wins either the first game or the second game, his score in this game must be at least 25 . This means that $d \geq 0$; since the scores are all distinct, it must be the case that $d>0$.
Since Viswanathan's scores in these games are both larger than 25, Magnus's scores in these games are either 2 more and 2 less than Viswanathan's or 2 less and 2 more than Viswanathan's, depending on the order in which they won these games.
Suppose that Viswanathan wins the first game and Magnus wins the second game.
Then Magnus's score in the first game is 2 less than Viswanathan's (and so is $23+2 d$ ) and Magnus's score in the second game is 2 more than Viswanathan's (and so is $27+d$ ).
\begin{tabular}{c|c|c|c|} 
& 1st game & 2nd game & 3rd game \\
\hline Viswanathan & $25+2 d$ & $25+d$ & 25 \\
\hline Magnus & $23+2 d$ & $27+d$ & \\
\hline
\end{tabular}

Since scores of $a, b, c$ in order of the three games gives $2 b=a+c$, then $c=2 b-a$. Here, this means that Magnus's score in the third game is $2(27+d)-(23+2 d)=31$, which is not possible since a final score of 31 to 25 is not possible.
Suppose that Magnus wins the first game and Viswanathan wins the second game.
Then Magnus's score in the first game is 2 more than Viswanathan's (and so is $27+2 d$ ) and Magnus's score in the second game is 2 less than Viswanathan's (and so is $23+d$ ).
\begin{tabular}{c|c|c|c|} 
& 1st game & 2nd game & 3rd game \\
\hline Viswanathan & $25+2 d$ & $25+d$ & 25 \\
\hline Magnus & $27+2 d$ & $23+d$ & \\
\hline
\end{tabular}

Here, Magnus's score in the third game is $2(23+d)-(27+2 d)=19$, which is possible since a score of 25 to 19 is possible.
Therefore, Magnus's score in the third game is 19. We note that this answer does not depend on $d$, which tells us that this score will be 19 regardless of the value of $d>0$.
ANSWER: 19