12. (6 points) There are 80 beads. Five years ago, two sisters divided them according to their age ratio, and they were exactly divided; this year, they divided them again according to their age ratio, and they were exactly divided again. It is known that the elder sister is 2 years older than the younger sister. Therefore, the number of beads the elder sister received in the two distributions differs by $\qquad$ beads.
【Solution】Solution: Let the sister's age 5 years ago be $x$, then:
\begin{tabular}{|c|c|c|}
\hline & 5 years ago & This year \\
\hline Sister & $x$ & $x+5$ \\
\hline Elder sister & $x+2$ & $x+7$ \\
\hline
\end{tabular}
5 years ago and this year, the ages are distributed in proportion, and are exactly divided, so $2x+2$ and $2x+12$ are both factors of 80, and the difference between these two factors is $10; 80$'s factors are $1,2,4,5,8,10,16,20,40,80$, so only 10 and 20 have a difference of 10, thus $2x+2=10$, solving for $x=4$.

Then $x+2=4+2=6$, meaning 5 years ago the distribution was in the ratio of 4:6, the elder sister received:
$$
\begin{array}{l}
80 \div(4+6) \times 6 \\
=80 \div 10 \times 6 \\
=48 \text { (pieces); }
\end{array}
$$
$x+5=9, x+7=11$, meaning this year the distribution is in the ratio of $9:11$, the elder sister received:
$$
\begin{array}{l}
80 \div(9+11) \times 11 \\
=80 \div 20 \times 11 \\
=4 \times 11
\end{array}
$$
$=44$ (pieces);

The difference between the two distributions:
$48-44=4$ (pieces).
Answer: The elder sister received 4 more pieces of beads in the two distributions.
The answer is: 4.