PROBLEM 2. Let the numbers $x=1+2+2^{2}+\ldots+2^{2014}$ and $y=\left(27^{3}: 3^{8}+10^{10}: 10^{8}-72\right)^{403}-2015^{0}$.

a) Show that $x=2^{2015}-1$;

b) Compare the numbers $x$ and $y$;

c) Find the last digit of the number $x+y$.
## PROBLEM 2.

$(1 p)$
a) $x=2 x-x=\left(2+2^{2}+2^{3}+\ldots+2^{2014}+2^{2015}\right)-\left(1+2+2^{2}+\ldots+2^{2014}\right)=2^{2015}-1$;

$(1 p)$

b) We have $y=31^{403}-1$

(1p) and $x=32^{403}-1$

(1p) therefore $x>y$;

(1p) c) Since $x$ is of the form $2^{4 k+3}-1(k \in \mathbb{N})$, the last digit of $x$ is $8-1=7$

(1p) The last digit of $y$ is $1-1=0$

(1p) Therefore, the last digit of $x+y$ is 7