## Subject 1.

a. Show that $13^{254}-13^{253}-13^{252}-13^{251}$ is divisible by 2014 for any natural number $n$.

b. Let the number $a=10^{448}+10^{224}-4$. Find how many digits the number $a$ has and calculate the sum of the digits of the number $a$.
## Subject 1.

a. More generally: $13^{n+3}-13^{n+2}-13^{n+1}-13^{n}=13^{n}\left(13^{3}-13^{2}-13-1\right)=13^{n} \cdot 2014: 2014 \quad \ldots 3 \mathbf{p}$.
b. $a=10^{448}+10^{224}-4=1 \underbrace{000 \ldots 0}_{448}+1 \underbrace{000 \ldots 0}_{224}-4=1 \underbrace{000 \ldots 0}_{224} \underbrace{999 \ldots}_{223} 96$

$a$ has 449 digits

$S=1+\underbrace{9+9+9+\ldots+9}_{223}+6=2014$