4. Divisibility of numbers. The number

$$
3^{105}+4^{105}
$$

is divisible by $13, 49, 181$ and 379, but not divisible by 5 and 11.

How to check this?
4. The expression $a^{n}+b^{n}$ is divisible by $a+b$ if $n$ is an odd number. Therefore, the number

$$
3^{105}+4^{105}=\left(3^{3}\right)^{35}+\left(4^{3}\right)^{35}
$$

is divisible by $3^{3}+4^{3}=7 \cdot 13$. Similarly, from the equalities

$$
\begin{aligned}
& 3^{105}+4^{105}=\left(3^{5}\right)^{21}+\left(4^{5}\right)^{21} \\
& 3^{105}+4^{105}=\left(3^{?}\right)^{10}+\left(4^{25}\right)^{15}
\end{aligned}
$$

it follows that this number is divisible by $35+45=7 \cdot 181$ and by $3 ?+$ $+4 ?=49.379$

Notice further that

$$
4^{3} \equiv-1(\bmod 5)
$$

(this formula means that the number $4^{3}$, when divided by 5, gives a remainder of -1). From this, it follows that

$$
\mathbf{4}^{105} \equiv(-1)^{35}(\bmod 5)
$$

thus,

$$
\mathbf{4}^{105} \equiv-1(\bmod 5)
$$

Similarly, $3^{2} \equiv-1(\bmod 5)$, from which it follows that

$$
3^{104} \equiv(-1)^{52}(\bmod 5)
$$

thus,

$$
3^{104} \equiv 1(\bmod 5)
$$

and

$$
3^{105} \equiv 3(\bmod 5)
$$

Since

$$
4^{105} \equiv-1(\bmod 5) \text { and } 3^{105} \equiv 3(\bmod 5)
$$

then

$$
3^{105}+4^{105} \equiv 2(\bmod 5)
$$

this means that the number $3^{105}+4^{105}$, when divided by 5, gives a remainder of 2.

Similarly,

$$
4^{3} \equiv-2(\bmod 11), \text { from which } 4^{15} \equiv-32(\bmod 11)
$$

and since $-32 \equiv 1(\bmod 11)$, then

$$
4^{15} \equiv 1(\bmod 11)
$$

and, ultimately,

$$
4^{105} \equiv 1(\bmod 11)
$$

Similarly, we verify that

$$
3^{5} \equiv 1(\bmod 11), \text { from which } 3^{105} \equiv 1(\bmod 11)
$$

Therefore,

$$
3^{105}+4^{105} \equiv 2 \quad(\bmod 11)
$$

this means that the number $3^{105}+4^{105}$, when divided by 11, gives a remainder of 2.