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Date May 2012 Marks available 7 Reference code 12M.3dm.hl.TZ0.5
Level HL only Paper Paper 3 Discrete mathematics Time zone TZ0
Command term Evaluate Question number 5 Adapted from N/A

Question

Use the result 2003=6×333+5 and Fermat’s little theorem to show that 220034(mod7) .

[3]
a.

Find 22003(mod11) and 22003(mod13).

[3]
b.

Use the Chinese remainder theorem, or otherwise, to evaluate 22003(mod1001), noting that 1001=7×11×13.

[7]
c.

Markscheme

22003=25×(26)333     M1A1

32×1(mod7) by Fermat’s little theorem     A1

4(mod7)     AG

[3 marks]

a.

2003=3+10×200     (M1)

22003=23×(210)200(8×1(mod11))8(mod11)     A1

22003=211×(212)1667(mod13)     A1

[3 marks]

b.

form M1=10017=143; M2=100111=91; M3=100113=77     M1

solve 143x11(mod7)x1=5     M1A1

x2=4; x3=12     A1A1

x=4×143×5+8×91×4+7×77×12=12240228(mod1001)     M1A1

[7 marks]

c.

Examiners report

Many candidates were able to complete part (a) and then went on to part (b). Some candidates raced through part (c). Others, who attempted part (c) using the alternative strategy of repeatedly solving linear congruencies, were sometimes successful.

a.

Many candidates were able to complete part (a) and then went on to part (b). Some candidates raced through part (c). Others, who attempted part (c) using the alternative strategy of repeatedly solving linear congruencies, were sometimes successful.

b.

Many candidates were able to complete part (a) and then went on to part (b). Some candidates raced through part (c). Others, who attempted part (c) using the alternative strategy of repeatedly solving linear congruencies, were sometimes successful.

c.

Syllabus sections

Topic 10 - Option: Discrete mathematics » 10.4 » Modular arithmetic.
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