Date | None Specimen | Marks available | 5 | Reference code | SPNone.3dm.hl.TZ0.1 |
Level | HL only | Paper | Paper 3 Discrete mathematics | Time zone | TZ0 |
Command term | Find and Hence or otherwise | Question number | 1 | Adapted from | N/A |
Question
Use the Euclidean algorithm to find the greatest common divisor of 259 and 581.
[4]
a.
Hence, or otherwise, find the general solution to the diophantine equation 259x + 581y = 7 .
[5]
b.
Markscheme
581=2×259+63 M1A1
259=4×63+7 A1
63=9×7
the GCD is therefore 7 A1
[4 marks]
a.
consider
7=259−4×63 M1
=259−4×(581−2×259) A1
=259×9+581×(−4) A1
the general solution is therefore
x=9+83n; y=−4−37n where n∈Z M1A1
Notes: Accept solutions laid out in tabular form. Dividing the diophantine equation by 7 is an equally valid method.
[5 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.