Date | November 2020 | Marks available | 4 | Reference code | 20N.3.AHL.TZ0.Hdm_3 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Find | Question number | Hdm_3 | Adapted from | N/A |
Question
Write down the remainder when 142022 is divided by 7.
Use Fermat’s little theorem to find the remainder when 142022 is divided by 17.
Prove that a number in base 13 is divisible by 6 if, and only if, the sum of its digits is divisible by 6.
The base 13 number 1y93y25 is divisible by 6. Find the possible values of the digit y.
Markscheme
the remainder is 0 A1
[1 mark]
1416≡1(mod 17) (from Fermat’s little theorem) (A1)
142022=1416×126+6 (M1)
Note: Award M1 for a RHS exponent consistent with the correct use of Fermat’s little theorem.
142022≡146(mod 17) (≡15(mod 17)) A1
the remainder is 15 A1
[4 marks]
METHOD 1
let N=an13n+an-113n-1+…+a113+a0 M1
Note: The above M1 is independent of the A marks below.
13≡1(mod 6) A1
EITHER
13x≡1(mod 6) (for all x∈ℕ) A1
OR
N≡an1n+an-11n-1+…+a11+a0 (mod 6) (N≡an+an-1+…+a1+a0 (mod 6)) A1
THEN
so N≡0 (mod 6) if and only if an+an-1+…+a1+a0≡0 (mod 6) R1
so 6 N if and only if 6 (an+an-1+…+a1+a0) AG
METHOD 2
let N=an13n+an-113n-1+…+a113+a0 (M1)
N=(an+an-1+…+a1+a0)+(13-1)(an(13n-1+…+130)+an-1(13n-2+…+130)+…+a1130) M1A1
Note: Award M1 for attempting to express N in the form N=(an+an-1+…+a1+a0)+(13-1)M.
as 6 (13-1)M R1
so 6 N if and only if 6 (an+an-1+…+a1+a0) AG
[4 marks]
METHOD 1
the sum of the digits is 2y+20 (A1)
uses 2y+20=6k (or equivalent) to attempt to find a valid value of y (M1)
y=2, 5, 8, 11(B) A1A1
Note: Award A1 for y=2, 5, 8 and A1 for y=11(B).
METHOD 2
(1y93y25)13=1×136+y×135+9×134+3×133+y×132+2×131+5×130 (A1)
=371462y+5090480
attempts to find a valid value of y such that
371462y+5090480≡0(mod 6) (M1)
y=2, 5, 8, 11(B) A1A1
Note: Award A1 for y=2, 5, 8 and A1 for y=11(B).
[4 marks]