NZMO Round One

Solution:

Let $f(n) = 122^{n} - 102^{n} - 21^{n}$. We consider $f(n)$ in mod 101 and in mod 20 separately.

Consider $f(n)$ mod 101.

$$\begin{align*} f(n) &= 122^{n} - 102^{n} - 21^{n} \\ &\equiv 21^{n} - 1^{n} - 21^{n} \pmod{101} \\ &= -1 \end{align*}$$

Consider $f(n)$ mod 20.

$$\begin{align*} f(n) &= 122^{n} - 102^{n} - 21^{n} \\ &\equiv 2^{n} - 2^{n} - 1^{n} \pmod{20} \\ &= -1 \end{align*}$$

Therefore $f(n) \equiv -1$ both in mod 20 and in mod 101. Since 20 and 101 are relatively prime, this means $f(n) \equiv -1$ (mod 2020). As required. $\square$