Let's determine the remainders when these numbers

$$
65^{6 n}, \quad 65^{6 n+1}, \quad 65^{6 n+2}, \quad 65^{6 n+3}
$$

are divided by 9.
The given expressions can also be written as:

$$
(63+2)^{6 n},(63+2)^{6 n+1},(63+2)^{6 n+2},(63+2)^{6 n+3}
$$

If we expand these expressions, we see that only the last term is not divisible by 9; therefore, it is sufficient to examine the remainders of the last terms.

But

$$
2^{6 n}, \quad 2^{6 n+1}, \quad 2^{6 n+2}, \quad 2^{6 n+3}
$$

can also be written as:

$$
64^{n}, \quad 2 \cdot 64^{n}, \quad 4 \cdot 64^{n}, \quad 8 \cdot 64^{n}
$$

or

$$
(63+1)^{n}, \quad 2 \cdot(63+1)^{n}, \quad 4 \cdot(63+1)^{n}, \quad 8 \cdot(63+1)^{n}
$$

from which it is clear that the remainders are: $1, 2, 4$ and 8.

(Béla Bayer, Losoncz.)

The problem was also solved by: F. Aczél, I. Bartók, P. Beck, G. Bogdán, K. Burján, K. Czank, J. Demeter, E. Deutsch, I. Deutsch, J. Filkorn, J. Handler, T. Kerekes, D. König, B. Krausz, Gy. Krisztián, A. Kürth, L. Lázár, Gy. Lukhaub, Gy. Lupsa, M. Pintér, Gy. Póka, M. Riesz, M. Russo, J. Scharff, R. Scheuer, J. Schwarz, S. Simon, Á. Singer, K. Smodics, F. Stromfeld, Z. Szávay, H. Szmodics, G. Tőtössy, E. Tézner, A. Weisz, S. Wohlstein.