Berkeley Math Circle Monthly Contest 1

Solution:

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 $10 \equiv 1 \pmod{9}$, $x \equiv y \equiv d_{n-1} + d_{n-2} + \cdots + d_{1} + d_{0} \pmod{9}$, so $9 \mid x - y$ and therefore also $9 \mid (x - y)(x^{3} + x^{2} y + x y^{2} + y^{3}) = x^{4} - y^{4}$.

Since $10 \equiv -1 \pmod{11}$, $x \equiv d_{n-1} - d_{n-2} + \cdots \mp d_{1} \pm d_{0} \pmod{11}$ and $y \equiv d_{0} + d_{1} + \cdots \mp d_{n-2} \pm d_{n-1} \pmod{11}$, i.e. $x \equiv \pm y$. In either case $x^{4} \equiv y^{4} \pmod{11}$, 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 $n^{4} - f(n)^{4}$.