233*. Is the sum $19^{19}+69^{69}$ divisible by 44?
$\triangle$ Since $44=4 \cdot 11$, where the numbers 4 and 11 are coprime, it is sufficient to check whether this sum is divisible by 4 and 11.

Divisibility by 4 is simple: use the "plus-minus" trick, adding and subtracting 1. We get:

$$
19^{19}+69^{69}=\left(19^{19}+1\right)+\left(69^{69}-1\right)
$$

The number in each of the parentheses is divisible by 4, so the entire sum is divisible by 4.

But what about divisibility by 11? The same operation - adding and subtracting 1 - does not work here. If we don't come up with something better, we will have to raise 19 to the 19th power, 69 to the 69th power, add the resulting numbers, and divide the sum by 11.

Fortunately, there is a way to avoid such a difficult task, although it is not very simple. We use the same "plus-minus" trick, adding and subtracting not 1, but $69^{19}$. Why exactly this power? Because the sum $19^{19}+69^{69}$ is divisible by 88, and therefore by 11. We will have:

$$
\begin{gathered}
19^{19}+69^{69}=\left(19^{19}+69^{19}\right)+\left(69^{69}-69^{19}\right)= \\
=\left(19^{19}+69^{19}\right)+69^{19}\left(69^{50}-1\right)
\end{gathered}
$$

Is the difference $69^{50}-1$ divisible by 11? This difference is divisible by $69^{2}-1$, but the latter difference is not divisible by 11, as it factors into terms that do not divide by 11:

$$
69^{2}-1=(69+1)(69-1)=70 \cdot 68
$$

But the difference $69^{50}-1$ is divisible by $69^{5}-1$. Could this new difference be divisible by 11? Let's transform it, representing 69 as $66+3$, so that we can use the divisibility of 66 by 11:

$$
\begin{gathered}
69^{5}-1=(66+3)^{5}-1=(66+3)(66+3)(66+3)(66+3)(66+3)-1= \\
=66 k+3^{5}-1
\end{gathered}
$$

(If you multiply five sums, each equal to $66+3$, then all terms of the resulting expansion, except one equal to $3^{5}$, are divisible by 66; their sum is denoted by $66 k$, where $k$ is a natural number.)

The number

$$
3^{5}-1=243-1=242
$$

is divisible by 11. Then the difference $69^{5}-1$ is divisible by 11, and therefore the difference $69^{50}-1$ is also divisible by 11. Consequently, the given sum $19^{19}+69^{69}$ is divisible by 11.

Answer: divisible.