【Question 12】There are 5 natural numbers (equality is allowed), from which any 4 numbers are selected to find their sum, and the only possible sums are $44, 45, 46$, 47. Therefore, the original 5 natural numbers are $\qquad$.
Analysis:
Let the 5 natural numbers be $a, b, c, d, e$, and let $m=a+b+c+d+e$.
From any 4 of these numbers, the sums are $m-a, m-b, m-c, m-d, m-e$. When these 5 numbers are all different, there should be 5 different sums. However, the problem states that only 4 different sums can be obtained, indicating that two of these 5 numbers are equal. Assume $a>b>c>d$ (with $e$ being the repeated number),
then $\left\{\begin{array}{l}m-a=44 \\ m-b=45 \\ m-c=46 \quad \text { (where } x \text { can be } 44, 45, 46, 47 \text { ) } \\ m-d=47 \\ m-e=x\end{array}\right.$
Adding the 5 equations, we get $5 m-(a+b+c+d+e)=44+45+46+47+x$
which simplifies to $4 m=182+x$
Solving this, we find $x=46, m=57$
Therefore, $a=13, b=12, c=e=11, d=10$