2. If $60=a-b$ holds, then it follows that $a \geqslant 61$, so $b$ and $c$ can be at most $100-61-1=38$. Therefore, the difference between the numbers $b$ and $c$ is at most $38-1=37$. Thus, for the condition of the problem to be satisfied, the only possibility is that the difference between the numbers $a$ and $c$ is 38, i.e., since $a$ is clearly greater than $c$, we have $38=a-c$. Now substituting $b=a-60$ and $c=a-38$ into $a+b+c=100$ gives $3a=198$, i.e., $a=66, b=6$ and $c=28$.
2. Let $a, b$, and $c$ be natural numbers whose sum is equal to 100. Consider all differences between any two of these three numbers. If it is known that one of these differences is equal to 60 and one is equal to 38, determine the numbers $a, b$, and $c$.