# Problem 3. (3 points)

$4^{27000}-82$ is divisible by $3^n$. What is the greatest natural value that $n$ can take?
Answer: 5

Solution:
$4^{27000}=(1+3)^{27000}=1+27000 \cdot 3+\frac{27000 \cdot 26999 \cdot 3^{2}}{2}+\frac{27000 \cdot \ldots \cdot 26998 \cdot 3^{3}}{6}+$
$+\frac{27000 \cdot \ldots \cdot 26997 \cdot 3^{4}}{24}+\frac{27000 \cdot \ldots \cdot 26996 \cdot 3^{5}}{120} \ldots$

The last two terms listed are divisible by $3^{6}$, as are all other terms included in the ellipsis.

$$
1+27000 \cdot 3+\frac{27000 \cdot 26999 \cdot 3^{2}}{2}+\frac{27000 \cdot 26999 \cdot 26998 \cdot 3^{3}}{6}=1+81000+
$$

$$
+\frac{27000 \cdot 26999 \cdot 3^{2}+27000 \cdot 26999 \cdot 26998 \cdot 3^{2}}{2}
$$

$1+81000=1+81(1+999)=1+81+81 \cdot 999=1+82+81 \cdot 999$, which gives a remainder of 82 when divided by $3^{6}$, since the last term is divisible by $3^{6}$.

$\frac{27000 \cdot 26999 \cdot 3^{2}+27000 \cdot 26999 \cdot 26998 \cdot 3^{2}}{2}=\frac{27000 \cdot 26999 \cdot 3^{2} \cdot(1+26998)}{2}$, and this number is clearly divisible by $3^{5}$, but not by $3^{6}$. Therefore, $4^{27000}-82$ is also divisible by $3^{5}$, but not by $3^{6}$.