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Date November 2019 Marks available 2 Reference code 19N.3.AHL.TZ0.Hdm_2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term State Question number Hdm_2 Adapted from N/A

Question

In parts (b) and (c), (abc)n denotes the number abc written in base n, where nZ+. For example, (359)n=3n2+5n+9.

State Fermat’s little theorem.

[2]
a.i.

Find the remainder when 151207 is divided by 13.

[5]
a.ii.

Convert (7A2)16 to base 5, where (A)16=(10)10.

[4]
b.

Consider the equation (1251)n+(30)n=(504)n+(504)n.

Find the value of n.

[4]
c.

Markscheme

EITHER

apa(modp)         A1

where p is prime         A1

 

OR

ap11(modp)         A1

where p is prime and p does not divide a (or equivalent statement)         A1

 

[2 marks]

a.i.

15120721207(mod13)

2121(mod13)        (M1)(A1)

21207=(212)10027        (M1)

21207(27)11(mod13)        (M1)A1

the remainder is 11

Note: Award as above for using 15 instead of 2.

[5 marks]

a.ii.

(7A2)16=7×162+10×16+2           M1

=1954         A1

EITHER

5|1954_

       390r4

          78r0

          15r3           M1

            3r0

            0r3

OR

1954=3×54+0×53+3×52+0×51+4           M1

THEN

(7A2)16=(30304)5         A1

 

[4 marks]

b.

the equation can be written as

n3+2n2+5n+1+3n=2(5n2+4)           M1A1

n38n2+8n7=0           (M1)

Note: The (M1) is for an attempt to solve the original equation.

n=7         A1

[4 marks]

c.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 1—Number and algebra » SL 1.7—Laws of exponents and logs
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Topic 1—Number and algebra

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