## SUBJECT I

a) Verify if the number

$$
a=2^{20}\left\{11^{2}: 121+3\left[\left(3^{5} \cdot 2^{3}\right)^{10}:\left(6^{30} \cdot 3^{19}\right)-2014^{0}\right]\right\}-7
$$

is divisible by 7 and 5.

b) Find the remainder of the division of the number $b=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 2013+2015$ by 2014.
## SUBJECT I

| I a) $\left(3^{5} \cdot 2^{3}\right)^{10}=3^{50} \cdot 2^{30}$ | $\mathbf{0 . 5 p}$ |
| :---: | :---: |
| $6^{30} \cdot 3^{19}=3^{49} \cdot 2^{30}$ | $\mathbf{1 p}$ |
| $\left(3^{50} \cdot 2^{30}\right):\left(3^{49} \cdot 2^{30}\right)=3$ | $\mathbf{0 . 5 p}$ |
| $2^{20}\left[11^{2}: 121+3 \cdot(3-1)\right]-7=2^{20} \cdot 7-7$ | $\mathbf{0 . 5 p}$ |
| $7\left(2^{20}-1\right)$ is divisible by 7 | $0.5 \mathbf{p}$ |
| $u\left(2^{20}\right)=6$ | $\mathbf{0 . 5 p}$ |
| $u\left(2^{20}-1\right)=5, u(a)=5$ so the number $a$ is divisible by 5 | $\mathbf{0 . 5 p}$ |
| I b) $b=1 \cdot 2 \cdot 3 \cdot \ldots \cdot 1006 \cdot 1007 \cdot 1008 \cdot \ldots \cdot 2013+2014+1$ | $0.5 \mathbf{p}$ |
| $b=1 \cdot 3 \cdot 4 \cdot \ldots \cdot 1006 \cdot 1008 \cdot \ldots \cdot 2013 \cdot(2 \cdot 1007)+2014+1$ | $1 \mathbf{1 p}$ |
| $b=1 \cdot 3 \cdot 4 \cdot \ldots \cdot 1006 \cdot 1008 \cdot \ldots \cdot 2013 \cdot 2014+2014+1$ | $0.5 \mathbf{p}$ |
| $b=2014 \cdot(1 \cdot 3 \cdot 4 \cdot \ldots \cdot 1006 \cdot 1008 \cdot \ldots \cdot 2012 \cdot 2013+1)+1$ | $0.5 \mathrm{p}$ |
| From the division theorem with remainder, the remainder is 1 | $0.5 \mathbf{p}$ |
| TOTAL | $7 \mathbf{p}$ |

## MATHEMATICS OLYMPIAD  LOCAL STAGE  16.02.2014

GRADE 5

## GRADING SCALE