Denote by $f(n)$ the integer obtained by reversing the digits of a positive integer $n$. Find the greatest integer that is certain to divide $n^{4}-f(n)^{4}$ regardless of the choice of $n$.
The answer is 99 . Let $x=\overline{d_{n-1} \ldots d_{2} d_{1} d_{0}}$, i.e. $x=10^{n-1} d_{n-1}+\cdots+10^{2} d_{2}+10 d_{1}+d_{0}$. Then $y=f(x)=10^{n-1} d_{0}+10^{n-2} d_{1}+\cdots+10 d_{n-2}+d_{n-1}$.

Let us show separately that $9 \mid x^{4}-y^{4}$ and that $11 \mid x^{4}-y^{4}$. Since $10 \equiv 1(\bmod 9), x \equiv y \equiv d_{n-1}+d_{n-2}+\cdots+d_{1}+d_{0}(\bmod 9)$, so $9 \mid x-y$ and therefore also $9 \mid(x-y)\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)=x^{4}-y^{4}$. Since $10 \equiv-1(\bmod 11), x \equiv d_{n-1}-d_{n-2}+$ $\cdots \mp d_{1} \pm d_{0}(\bmod 11)$ and $y \equiv d_{0}+d_{1}+\cdots \mp d_{n-2} \pm d_{n-1}(\bmod 11)$, i.e. $x \equiv \pm y$. In either case $x^{4} \equiv y^{4} \bmod 11$, so $11 \mid x^{4}-y^{4}$. Since 9 and 11 are relatively prime, their product 99 must divide $x^{4}-y^{4}$.

To see that there is no larger integer that always divides $n^{4}-f(n)^{4}$, let $n=10$, so that $f(n)=01=1, n^{4}-f(n)^{4}=9999=$ $3^{2} \cdot 11 \cdot 101$, and let $n=21$, so that $f(n)=12$ and $n^{4}-f(n)^{4}=173745=3^{5} \cdot 5 \cdot 11 \cdot 13$. The greatest common divisor of these two values of $n^{4}-f(n)^{4}$ is $3^{2} \cdot 11=99$, so there is no larger integer certain to divide $n^{4}-f(n)^{4}$.

Remark. From the solution it is easy to see that 99 is even the common divisor of the numbers of the form $n^{2}-f(n)^{2}$.