## Task A-4.3. (4 points)

What is the remainder when the number (ABCDE2010) ${ }_{15}$ is divided by 7?

Numbers in base 15 are written using the digits $0,1,2,3,4,5,6,7,8,9$, A, B, C, D, E whose values are respectively $0,1,2,3,4,5,6,7,8,9,10,11,12,13,14$.
## First Solution.

The number (ABCDE2010) ${ }_{15}$ can be written as

$$
a=10 \cdot 15^{8}+11 \cdot 15^{7}+12 \cdot 15^{6}+13 \cdot 15^{5}+14 \cdot 15^{4}+2 \cdot 15^{3}+1 \cdot 15^{1}
$$

Let $b=10+11+12+13+14+2+0+1+0=63$.

Notice that the number $15^{k}-1$ is divisible by 7 for any natural number $k$ because

$$
15^{k}-1=(15-1) \cdot\left(15^{k-1}+\ldots+1\right)
$$

Subtract the number $b$ from the number $a$:

$$
\begin{aligned}
& a-b= \\
& =10 \cdot 15^{8}+11 \cdot 15^{7}+12 \cdot 15^{6}+13 \cdot 15^{5}+14 \cdot 15^{4}+2 \cdot 15^{3}+1 \cdot 15^{1} \\
& \begin{array}{lllllll}
-10 & -11 & -12 & -13 & -14 & -2 & -1
\end{array} \\
& =10 \cdot\left(15^{8}-1\right)+11 \cdot\left(15^{7}-1\right)+12 \cdot\left(15^{6}-1\right)+13 \cdot\left(15^{5}-1\right)+14 \cdot\left(15^{4}-1\right)+ \\
& +2 \cdot\left(15^{3}-1\right)+1 \cdot(15-1)
\end{aligned}
$$

We conclude that this difference is divisible by 7 because each addend is divisible by 7.

(2 points)

On the other hand, we know that the number $b=63$ is divisible by 7, so it follows that the number $a$ is also divisible by 7.

Therefore, the remainder when divided by 7 is 0.

Note. The fact that when dividing $15^{k}$ by 7 we get a remainder of 1 (for $k=1, \ldots, 8$) can also be obtained using the binomial formula

$$
15^{k}=(14+1)^{k}=\sum_{i=0}^{k}\binom{k}{i} 14^{i}
$$

All terms in the sum are divisible by 7 except the first $\binom{k}{0} \cdot 1=1$.