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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=2594×63     M1

=2594×(5812×259)     A1

=259×9+581×(4)     A1

the general solution is therefore

x=9+83n; y=437n where nZ     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.

Syllabus sections

Topic 10 - Option: Discrete mathematics » 10.3 » Linear Diophantine equations ax+by=c .

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