Asian Pacific Mathematics Olympiad

In both items, the smallest chosen number is at least $2$: in part (a), $n / 2 > 1$ and in part (b), $p$ is a prime. So let $1 < x < y < z$ be the chosen numbers. Then all possible combinations are
$$x + y + z, \quad x + y z, \quad x y + z, \quad y + z x, \quad (x + y) z, \quad (z + x) y, \quad (x + y) z, \quad x y z.$$

Since, for $1 < m < n$ and $t > 1$, $(m-1)(n-1) \geq 1 \cdot 2 \Longrightarrow m n > m + n$, $t n + m - (t m + n) = (t-1)(n-m) > 0 \Longrightarrow t n + m > t m + n$, and $(t + m) n - (t + n) m = t(n-m) > 0$,
$$x + y + z < z + x y < y + z x < x + y z$$

and
$$(y + z) x < (x + z) y < (x + y) z < x y z.$$

Also, $(y + z) x - (y + z x) = (x - 1) y > 0 \Longrightarrow (y + z) x > y + z x$ and $(x + z) y - (x + y z) = (y - 1) x > 0 \Longrightarrow (x + z) y > x + y z$. Therefore the only numbers that can be equal are $x + y z$ and $(y + z) x$. In this case,
$$x + y z = (y + z) x \Longleftrightarrow (y - x)(z - x) = x(x - 1).$$

Now we can solve the items.

a. If $n / 2 < x < y < z$ then $z - x < n / 2$, and since $y - x < z - x$, $y - x < n / 2 - 1$; then
$$(y - x)(z - x) < \frac{n}{2} \left( \frac{n}{2} - 1 \right) < x(x - 1),$$

and therefore $x + y z < (y + z) x$.

b. If $x = p$, then $(y - p)(z - p) = p(p - 1)$. Since $y - p < z - p$, $(y - p)^2 < (y - p)(z - p) = p(p - 1) \Longrightarrow y - p < p$, that is, $p$ does not divide $y - p$. Then $y - p$ is a divisor $d$ of $p - 1$ and $z - p = \frac{p(p - 1)}{d}$. Therefore,
$$x = p, \quad y = p + d, \quad z = p + \frac{p(p - 1)}{d},$$
which is a solution for every divisor $d$ of $p - 1$ because
$$x = p < y = p + d < 2p \leq p + p \cdot \frac{p - 1}{d} = z.$$