57. The number $123456789(10)(11)(12)(13)(14)$ is written in the base-15 numeral system, i.e., this number is equal to

(14) $+(13) \cdot 15+(12) \cdot 15^{2}+(11) \cdot 15^{3}+\ldots+2 \cdot 15^{12}+15^{13}$. What remainder does it give when divided by 7?
57. The number 15 gives a remainder of 1 when divided by 7. Therefore, it follows that

$$
15^{2}=(7 \cdot 2+1) \cdot(7 \cdot 2+1)=7 n_{1}+1
$$

gives a remainder of 1 when divided by 7,

$$
15^{3}=15^{2} \cdot 15=\left(7 n_{1}+1\right) \cdot(7 \cdot 2+1)=7 n_{2}+1
$$

gives a remainder of 1 when divided by 7, and generally, any power of the number 15 gives a remainder of 1 when divided by 7. If we now subtract from the given number the sum \(1+2+3+4+\ldots+14=105\), then, by grouping the terms appropriately, we get:

$$
\begin{aligned}
& 13(15-1)+12\left(15^{2}-1\right)+ \\
& \quad+11\left(15^{3}-1\right)+\ldots+2\left(15^{12}-1\right)+1\left(15^{13}-1\right)
\end{aligned}
$$

i.e., a number that is divisible by 7. But from the fact that the difference between the given number and the number \(105=7 \cdot 15\) is divisible by 7, it follows that the original number is also divisible by 7.