A game is played with four integers in the following way: add three of these numbers, divide this sum by 3, and then add the result to the fourth number. There are four ways to do this game, yielding the following results: 17, 21, 23, and 29. What is the largest of the four numbers?
Let $a, b, c$ and $d$ be the numbers we are looking for. The given numbers are

$$
\frac{a+b+c}{3}+d, \frac{a+b+d}{3}+c, \frac{a+c+d}{3}+b \quad \text { and } \quad \frac{b+c+d}{3}+a
$$

but we do not know their order. However,

$$
\begin{aligned}
90 & =17+21+23+29 \\
& =\frac{a+b+c}{3}+d+\frac{a+b+d}{3}+c+\frac{a+c+d}{3}+b+\frac{b+c+d}{3}+a \\
& =2(a+b+c+d)
\end{aligned}
$$

and therefore, $2(a+b+c+d)=90$, or $a+b+c+d=45$. Now, let $d$ be the largest of the numbers $a, b, c$ and $d$. Then

$$
d=29-\frac{a+b+c}{3}=29-\frac{45-d}{3}
$$

and we conclude that $d=21$.