1. Let the numbers be $m=(1+2+3)^{4}+5 \cdot 6 \cdot(7+8+9)$ and $n=1+2+3+\ldots+63$.

a) (3p) Calculate $2016^{m-n}$.

b) (4p) Determine the largest and the smallest non-zero natural number which, when divided by $n$, gives a quotient equal to a quarter of the remainder.

Prof. Stela Boghian
## Solution:

a) $m=6^{4}+5 \cdot 6 \cdot 24=1296+720=2016$ and $n=63 \cdot 64: 2=2016 \Rightarrow 2016^{m-n}=2016^{0}=1$.

b) From the division theorem with remainder, we have $d=2016 \cdot c+r ; r$ must be a multiple of 4. | $1 p$ |
|  | The largest number is obtained for $r=2012$, which is 1016060. | $2 p$ |
|  | and the smallest non-zero number for $r=4$; this is 2020. | $1 p$ |