What is the remainder when 74 to the 600th power is divided by 9?
$$
74^{600}=(72+2)^{600}=72^{600}+\binom{600}{1} 72^{599} \cdot 2+\ldots+\binom{600}{599} 72 \cdot 2^{599}+2^{600}
$$

Except for the last term, every term is divisible by 9.

But the last term

$$
2^{600}=\left(2^{6}\right)^{100}=(63+1)^{100}
$$

Again, by the binomial theorem, it is obvious that only the last term \(1^{100}=1\) is not divisible by 9.

Therefore, the sought remainder is 1.

Sandor Holbok (Makó, S. Kovács János g. III. o. t.)