2. Determine five natural numbers $a, b, c, d, e$ (greater than 0), such that

$$
a+b+c+d+e=100, \quad \operatorname{LCM}(a, b, c, d, e)=1992
$$

Without considering solutions that are obtained from another solution by permuting the numbers.
Solution. We have $1992=2^{3} \cdot 3 \cdot 83$, so at least one of the numbers $a, b, c, d$ must be divisible by 83, because otherwise 83 would not appear as a prime factor in their LCM. If two of the numbers are divisible by 83, then their sum would be greater than 100, which is not possible. Therefore, exactly one of the numbers $a, b, c, d, e$ is divisible by 83. Let's set $a=83k$. $k$ must be one, because otherwise $a>100$ which is not possible. Therefore, $a=83$. From $b+c+d+e=17$ and $LCM(83, a, b, c, d, e)=2^{3} \cdot 3 \cdot 83$ it follows that at least one of the numbers $b, c, d, e$ is divisible by $2^{3}=8$ because otherwise 2 would appear in the LCM $(83, a, b, c, d, e)$ at a lower power or not at all. If two of the numbers $b, c, d, e$ are divisible by 8, then their sum with $a$ is greater than or equal to 99, so for the remaining two numbers we would have that their sum is 1, which is not possible. Therefore, exactly one of the numbers $b, c, d, e$ is divisible by 8. Let's set $b=8m$. But $m$ must be one, because otherwise $a+b \geq 99$, i.e., $c+d+e \leq 1$ which is not possible. Therefore, $b=8$. From $c+d+e=9$ and $LCM(83,8, b, c, d, e)=2^{3} \cdot 3 \cdot 83$, it follows that at least one of the numbers $c, d$ and $e$ must be divisible by 3, because otherwise 3 would not appear as a factor in $LCM(83,8, b, c, d, e)=2^{3} \cdot 3 \cdot 83$. Let's set $c=3n$. If $c=3$, then $d+e=6$ and $d, e$ have possible divisors only $1,2,3$ and 4; from this, two solutions are obtained: $d=e=3$ and $d=4, e=2$. If $c=6$, then $d+e=3$ and $d, e$ have possible divisors $1,2,3$ and 4, from which the solution $d=2, e=1$ is obtained. If $c \geq 9$, then $d+e \leq 0$ which is not possible. Therefore, there are three solutions, and they are

$$
\begin{aligned}
& a=83 ; \quad b=8 ; \quad c=3 ; \quad d=3 ; \quad e=3 \\
& a=83 ; \quad b=8 ; \quad c=3 ; \quad d=4 ; \quad e=2 \\
& a=83 ; \quad b=8 ; \quad c=6 ; \quad d=2 ; \quad e=1
\end{aligned}
$$