A game is played with four integers in the following way: add three of these numbers, divide this sum by 3, and 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

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

but we do not know their order. Since

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

Thus,

$$
2(a+b+c+d)=90 \Longrightarrow a+b+c+d=45
$$

Now, let $d$ be the largest among the numbers $a, b, c$ and $d$. Therefore,

$$
d=29-\frac{a+b+c}{3}=29-\frac{45-d}{3} \Longrightarrow d=21
$$