## SUBJECT III

Let $m$ be a non-zero natural number. Consider the sets:

$A=\left\{x \in N / x=2^{n}-1, n \in N^{*}, n \leq m\right\}$ and $B=\left\{y \in N / y=2^{n-1}, n \in N^{*}, n \leq m\right\}$.

a) Explain why the set $A \cap B$ cannot have more than one element.

b) Determine the number $m$ knowing that the set $A \cup B$ contains only consecutive numbers.

c) Determine the minimum value of the number $m$ such that the statement " $1023 \in A$ " is true.

d) What is the maximum value of the number $m$ if the set $B$ contains exactly 8 perfect squares?
## SUBJECT III

$\mathrm{A}=\{1,3,7,15,31,63\}$ and $B=\{1,2,4,8,16,32\}$

a) except for 1, B contains only even numbers, while A contains only odd numbers. The only common element can be 1. $1 \mathrm{p}$ Note: If the student does not fully explain but only writes that for $\mathrm{m}=1 \Rightarrow \mathrm{n}=1 \Rightarrow A \cap B=\{1\}$, then $0.5 \mathrm{p}$ is awarded.

b) If $m=2 \Rightarrow n \in\{1,2\} \Rightarrow A=\{1,3\}$ and $B=\{1,2\}$, then $A \cup B=\{1,2,3\}$, which is the only set with consecutive numbers. $2 \mathrm{p}$
c) $1023 \in \mathrm{A}$, so $2^{\mathrm{n}}-1=1023, \mathrm{n}=10$ thus $\mathrm{m}=10$. $\qquad$
d) $\mathrm{y}=2^{\mathrm{n}-1}$ is a perfect power (p.p.) if $\mathrm{n}-1$ is even.

The first 8 perfect powers are obtained for the first 8 even exponents, i.e., $n-1 \in\{0,2,4,6,8,10,12,14\}, n \in\{1,3,5,7,9,11,13,15\}$

The maximum value of m is 16.

$2 \mathrm{p}$

Note: If a student considers $\mathrm{m}=15$ instead of $\mathrm{m}=16$, then $1.75 \mathrm{p}$ is awarded.

Grading: a) $1 \mathrm{p}$
b) $2 \mathrm{p}$
c) $2 \mathrm{p}$
d) $2 \mathrm{p}$