Show that there are exactly two solutions to the equation $1982 = 36a + 74b$, with $a,{\text{ }}b \in \mathbb{N}$.

Hence, or otherwise, find the remainder when ${1982^{1982}}$ is divided by $37$.

$74 = 2 \times 36 + 2$ OR $\gcd (36,{\text{ }}74) = 2$     (A1)

$2 = ( - 2)(36) + (1)(74)$     M1

$1982 = ( - 1982)(36) + (991)(74)$     A1

so solutions are $a =  - 1982 + 37t$ and $b = 991 - 18t$     M1A1

$a,{\text{ }}b \in \mathbb{N}{\text{ so }}\frac{{1982}}{{37}} \le t \le \frac{{991}}{{18}}\;\;\;{\text{(}}15.56 \ldots  \le t \le 55.055 \ldots )$     (M1)(A1)

$t$ can take values $54$ or $55$ only     A1AG

(Or the solutions are $(16,{\text{ }}19)$ or $(53,{\text{ }}1)$)

Note:     Accept arguments from one solution of increasing/decreasing $a$ by $37$ and increasing/decreasing $b$ by $18$ to give the only possible positive solutions.

[8 marks]

$1982 = 53 \times 36 + 74$

$1982 = 55 \times 36 + 2$     (M1)

${1982^{36}} \equiv 1(\bmod 37){\text{ }}({\text{by FLT}})$     (M1)

${1982^{1982}} = {1982^{36 \times 55 + 2}} \equiv {1982^2}(\bmod 37)$     (A1)

$\equiv 34(\bmod 37)$     A1

so the remainder is $34$     A1

Note:     $1982$ in the base can be replaced by $21$.

Note:     Award the first (M1) if the $1982$ in the experiment is correctly broken down to a smaller number.

[5 marks]

Total [13 marks]