User interface language: English | Español

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),  ( a b c ) n  denotes the number  a b c  written in base n , where  n Z + . For example,  ( 359 ) n = 3 n 2 + 5 n + 9 .

State Fermat’s little theorem.

[2]
a.i.

Find the remainder when  15 1207 is divided by  13 .

[5]
a.ii.

Convert  ( 7 A 2 ) 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

a p a ( mod p )          A1

where p is prime         A1

 

OR

a p 1 1 ( mod p )          A1

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

 

[2 marks]

a.i.

15 1207 2 1207 ( mod 13 )

2 12 1 ( mod 13 )         (M1)(A1)

2 1207 = ( 2 12 ) 100 2 7         (M1)

2 1207 ( 2 7 ) 11 ( mod 13 )         (M1)A1

the remainder is 11

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

[5 marks]

a.ii.

( 7 A 2 ) 16 = 7 × 16 2 + 10 × 16 + 2            M1

= 1954          A1

EITHER

5 | 1954 _

        390 r 4

          78 r 0

          15 r 3            M1

            3 r 0

            0 r 3

OR

1954 = 3 × 5 4 + 0 × 5 3 + 3 × 5 2 + 0 × 5 1 + 4            M1

THEN

( 7 A 2 ) 16 = ( 30304 ) 5          A1

 

[4 marks]

b.

the equation can be written as

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

n 3 8 n 2 + 8 n 7 = 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
Show 49 related questions
Topic 1—Number and algebra

View options