## Subject 1.

a. Determine the natural number $n$ knowing that the sum of the digits of the number $a=10^{2 n}+10^{n}-2$ is equal to 2016.

b. Prove that $2^{2016}-2^{2010}$ is divisible by 2016.
## Subject 1.

| a. | $a=1 \underbrace{00 \ldots 00}_{n-1 \text { zeros }} 1 \underbrace{00 \ldots 00}_{\text {zeros }}-2=1 \underbrace{00 \ldots 00}_{\text {zeros }} \underbrace{99 \ldots 99}_{n-1 \text { digits of } 9} 8$  Sum of digits: $9(n-1)+8+1=2016$  Finalization $n=224$ | $\mathbf{2 p}$  $1 p$  $1 p$ |
| :---: | :---: | :---: |
| b. | $a=2^{2000}\left(2^{6}-1\right)=2^{2000} \cdot 63$  $2016=63 \cdot 32$  $a=2^{1995} \cdot 2^{5} \cdot 63=2^{1995} \cdot 2016$, so it is divisible by 2016 | $1 \mathrm{p}$  $\mathbf{1 p}$  $1 \mathrm{p}$ |