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]