Answer. There are ten triples satisfying the three conditions. They are given by $(1, 1, 1)$, $(1, 1, 2)$, $(1, 3, 2)$, $(3, 5, 4)$ and their cyclic permutations.

Without loss of generality, let $x$ be the smallest of the three numbers (or one of the smallest), i.e. $x \le y$ and $x \le z$. From $z \mid x + 1$ we obtain $x \le z \le x + 1$. Thus we have to consider two cases.

* Case 1. Let $z = x$. Then $z = x \mid x + 1$ leads to $x = z = 1$ and $y \mid z + 1 = 2$. Therefore $y = 1$ or $y = 2$, and we get the two solutions $(1, 1, 1)$ and $(1, 2, 1)$.

* Case 2. Let $z = x + 1$. Then the two conditions $x \mid y + 1$ and $y \mid x + 2$ must be fulfilled. In particular, we obtain $x \le y + 1$ and $y \le x + 2$. This yields $x - 1 \le y \le x + 2$ and we have to examine the following cases for $y$.

* Case 2a. Let $0 < y = x$. The conditions $x \mid x + 1$ and $x \mid x + 2$ can only hold simultaneously for $x = 1$, giving the solution $(1, 1, 2)$.

* Case 2b. Let $y = x + 1$. Then the two conditions are $x \mid x + 2$ and $x + 1 \mid x + 2$. They cannot hold simultaneously.

* Case 2c. Let $y = x + 2$. The condition $y = x + 2 \mid x + 2 = z + 1$ is trivially fulfilled. The requirement $x \mid y + 1 = x + 3$ can only hold for $x \mid 3$. And, indeed, for either $x = 1$ or $x = 3$ the condition is fulfilled and we obtain the solutions $(1, 3, 2)$ and $(3, 5, 4)$.

Summing up, the triples $(1, 1, 1)$, $(1, 2, 1)$, $(1, 1, 2)$, $(1, 3, 2)$ and $(3, 5, 4)$ fulfill all three conditions. As each of the three numbers can be the minimum, every cyclic permutation of these triples is a solution as well.