2. Determine all positive integers that are equal to 300 times the sum of their digits.
SOLUTION: We will prove that there is a unique solution, namely $n=2700$.

Let $n$ be a positive integer that satisfies the given conditions. We immediately observe that, since $n$ is a multiple of 300, and thus of 100, the units and tens digits of $n$ must be zero. Suppose, then, that its decimal representation is $n=a_{k} a_{k-1} \cdots a_{2} 00$, where, as usual, the leading digit $a_{k}$ is not zero. We have

$$
n=10^{k} a_{k}+10^{k-1} a_{k-1}+\cdots+100 a_{2} \geq 10^{k}
$$

and, since all digits are less than or equal to 9,

$$
s:=a_{k}+a_{k-1}+\cdots+a_{2} \leq 9(k-1).
$$

From the hypothesis $n=300 s$ we have, by the previous inequalities, $10^{k} \leq 300 \cdot 9(k-1)=2700(k-1)$. We verify by induction that this last inequality is false for $k \geq 4$: the base case is the verification that $10^{4}>2700 \cdot 3=8100$; for the inductive step, suppose that $10^{k}>2700(k-1)$: then $10^{k+1}>10^{k} \cdot 2>2700(2 k-2)>2700(k-2)$.

It follows that $k \leq 3$. Therefore, $n$ is of the form $10^{3} a_{3}+10^{2} a_{2}$ (in this case, it would a priori be allowed for $a_{3}=0$, although as we will see, this does not occur). The equation $n=300 s$ becomes $1000 a_{3}+100 a_{2}=300 a_{3}+300 a_{2}$, which simplifies to

$$
7 a_{3}=2 a_{2}.
$$

From this last equation, we deduce that 7 is a divisor of $a_{2}$, so the only possibility for a positive integer is $a_{2}=7$ and consequently $a_{3}=2$. On the other hand, the number thus found, $n=2700$, indeed satisfies the conditions of the problem.