Use the Euclidean algorithm to find $\gcd (162,{\text{ }}5982)$.

The relation $R$ is defined on ${\mathbb{Z}^ + }$ by $nRm$ if and only if $\gcd (n,{\text{ }}m) = 2$.

(i)     By finding counterexamples show that $R$ is neither reflexive nor transitive.

(ii)     Write down the set of solutions of $nR6$.

$5982 = 162 \times 36 + 150$    M1A1

$162 = 150 \times 1 + 12$    A1

$150 = 12 \times 12 + 6$

$12 = 6 \times 2 + 0 \Rightarrow \gcd$ is 6     A1

[4 marks]

(i)     for example, $\gcd (4,{\text{ }}4) = 4$     A1

$4 \ne 2$    R1

so $R$ is not reflexive     AG

for example

$\gcd (4,{\text{ }}2) = 2$ and $\gcd (2,{\text{ }}8) = 2$     M1A1

but $\gcd (4,{\text{ }}8) = 4{\text{ }}( \ne 2)$     R1

so $R$ is not transitive     AG

(ii)     EITHER

even numbers     A1

not divisible by 6     A1

OR

$\{ 2 + 6n:n \in \mathbb{N}\} {\text{ }} \cup \{ 4 + 6n:n \in \mathbb{N}\}$   A1A1

OR

$2,{\text{ }}4,{\text{ }}8,{\text{ }}10,{\text{ }} \ldots$    A2

[7 marks]

This was a successful question for many students with many wholly correct answers seen. Part (a) was successfully answered by most candidates and those candidates usually had a reasonable understanding of how to complete part (b). A number were not fully successful in knowing how to explain their results.

This was a successful question for many students with many wholly correct answers seen. Part (a) was successfully answered by most candidates and those candidates usually had a reasonable understanding of how to complete part (b). A number were not fully successful in knowing how to explain their results.