23 There are 88 numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{88}$ and each of them is either equals to -3 or -1 . Given that $a_{1}{ }^{2}+a_{2}{ }^{2}+\ldots+a_{88}{ }^{2}=280$, find the value of $a_{1}{ }^{4}+a_{2}{ }^{4}+\ldots+a_{88}{ }^{4}$.
23. Answer: 2008
Let $m$ of them be -3 and $n$ of them be -1 . Then $m+n=88$ and $(-3)^{2} m+(-1)^{2} n$ $=280$. Solving, $m=24, n=64$. Hence $(-3)^{4} m+(-1)^{4} n=2008$.