jmc

By the Binomial Theorem, $$\begin{array}{rl}(n-2{)}^5& =n^5-\left(\genfrac{}{}{0ex}{}{5}{1}\right)\cdot 2n^4+\left(\genfrac{}{}{0ex}{}{5}{2}\right)\cdot 2^2{n}^3-\left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 2^3{n}^2\\ & +\left(\genfrac{}{}{0ex}{}{5}{4}\right)\cdot 2^{4}n-2^5\\ & =n^5-10n^4+40n^3-80n^2+80n-32.\end{array}$$ Note that this reduces to $n^5-32\equiv n^5+3(\mathrm{mod}5)$. Therefore, $$\begin{array}{rl}8(n-2{)}^5-n^2+14n-24& \equiv 8(n^5+3)-n^2+14n-24\\ & \equiv 8n^5+24-n^2+14n-24\\ & \equiv 3n^5-n^2-n(\mathrm{mod}5).\end{array}$$

If $n\equiv 0(\mathrm{mod}5)$, then $$3n^5-n^2-n\equiv 3\cdot 0^5-0^2-0\equiv 0(\mathrm{mod}5).$$ If $n\equiv 1(\mathrm{mod}5)$, then $$3n^5-n^2-n\equiv 3\cdot 1^5-1^2-1\equiv 1(\mathrm{mod}5).$$ If $n\equiv 2(\mathrm{mod}5)$, then $$3n^5-n^2-n\equiv 3\cdot 2^5-2^2-2\equiv 90\equiv 0(\mathrm{mod}5).$$ If $n\equiv 3(\mathrm{mod}5)$, then $$3n^5-n^2-n\equiv 3\cdot 3^5-3^2-3\equiv 717\equiv 2(\mathrm{mod}5).$$ If $n\equiv 4(\mathrm{mod}5)$, then $$3n^5-n^2-n\equiv 3\cdot 4^5-4^2-4\equiv 3052\equiv 2(\mathrm{mod}5).$$

Therefore, the given expression is a multiple of 5 if and only if $n\equiv 0$ or $n\equiv 2(\mathrm{mod}5)$.

The largest value of $n$ less than 100000 that is congruent to 0 or 2 modulo 5 is $\boxed{99997}$.