Using the Euclidean algorithm, find h .

Hence find integers A and B such that 861A + 957B = h .

Using part (b), solve \(287w \equiv 2(\bmod 319\)) , where \(w \in \mathbb{N},{\text{ }}w < 319\) .

Find the general solution to the diophantine equation \(861x + 957y = 6\) .

METHOD 1

by the above working on the left (or similar)     M1A1A1

Note: Award A1 for 96 and A1 for 93.

h = 3 (since 3 divides 93)     A1

[4 marks]

METHOD 2

\(957 = 861 + 96\)     M1A1

\(861 = 8 \times 96 + 93\)     A1

\(96 = 93 + 3\)

so h = 3 (since 3 divides 93)     A1

[4 marks]

METHOD 1

if method 1 was used for part (a)

by the above working on the right (or equivalent)     M1

\( - 10 \times 861 + 9 \times 957 = 3\)

so A = –10 and B = 9     A1A1

[3 marks]

METHOD 2

\(3 = 96 - 93\)

\( = 96 - (861 - 8 \times 96) = 9 \times 96 - 861\)     M1

\( = 9 \times (957 - 861) - 861\)

\( = - 10 \times 861 + 9 \times 957\)

so A = –10 and B = 9     A1A1

[3 marks]

from (b) \( - 10 \times 287 + 9 \times 319 = 1\) so     A1

\( - 10 \times 287 \equiv 1(\bmod 319)\)     A1

\(287w \equiv 2(\bmod 319) \Rightarrow - 10 \times 287w \equiv - 10 \times 2(\bmod 319)\)     M1

\( \Rightarrow w \equiv - 20(\bmod 319)\)     A1

so \(w = 299\)     A1

[5 marks]

from (b) \( - 10 \times 861 + 9 \times 957 = 3 \Rightarrow - 20 \times 861 + 18 \times 957 = 6\)     M1A1

so general solution is \(x = - 20 + 319t\)     A1A1

\(y = 18 - 287t\,\,\,\,\,(t \in \mathbb{Z})\)     A1A1

[6 marks]

The Euclidean algorithm was well applied. If it is done in the format shown in the mark scheme then the keeping track method of the linear combinations of the 2 original numbers makes part (b) easier.

Again well answered but not quite as good as (a).

Surprisingly, since it is basic bookwork, this part was answered very badly indeed. Most candidates did not realise that \( - 10\) was the number to multiply by. Sadly, of the candidates that did do it, some did not read the question carefully enough to see that a positive integer answer was required.

Again this is standard bookwork. It was answered better than part (c). There were the usual mistakes in the final answer e.g. not having the two numbers, with the parameter, co-prime.