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 n∈Z+. For example, (359)n=3n2+5n+9.
State Fermat’s little theorem.
Find the remainder when 151207 is divided by 13.
Convert (7A2)16 to base 5, where (A)16=(10)10.
Consider the equation (1251)n+(30)n=(504)n+(504)n.
Find the value of n.
Markscheme
EITHER
ap≡a(modp) A1
where p is prime A1
OR
ap−1≡1(modp) A1
where p is prime and p does not divide a (or equivalent statement) A1
[2 marks]
151207≡21207(mod13)
212≡1(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]
(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]
the equation can be written as
n3+2n2+5n+1+3n=2(5n2+4) M1A1
⇒n3−8n2+8n−7=0 (M1)
Note: The (M1) is for an attempt to solve the original equation.
n=7 A1
[4 marks]