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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		Question number	Hdm_3	Adapted from	N/A

Question

Write down the remainder when $14^{2022}$ is divided by $7$ .

[1]

a.i.

Use Fermat’s little theorem to find the remainder when $14^{2022}$ is divided by $17$ .

[4]

a.ii.

Prove that a number in base $13$ is divisible by $6$ if, and only if, the sum of its digits is divisible by $6$ .

[4]

b.i.

The base $13$ number $1 y 93 y 25$ is divisible by $6$ . Find the possible values of the digit $y$ .

[4]

b.ii.

Markscheme

the remainder is $0$ A1

[1 mark]

a.i.

$14^{16} \equiv 1 (mod 17)$ (from Fermat’s little theorem) (A1)

$14^{2022} = 14^{16 \times 126 + 6}$ (M1)

Note: Award M1 for a $RHS$ exponent consistent with the correct use of Fermat’s little theorem.

$14^{2022} \equiv 14^{6} (mod 17) (\equiv 15 (mod 17))$ A1

the remainder is $15$ A1

[4 marks]

a.ii.

METHOD 1

let $N = a_{n} 13^{n} + a_{n - 1} 13^{n - 1} + \dots + a_{1} 13 + a_{0}$ M1

Note: The above M1 is independent of the A marks below.

$13 \equiv 1 (mod 6)$ A1

EITHER

$13^{x} \equiv 1 (mod 6)$ (for all $x \in ℕ$ ) A1

OR

$N \equiv a_{n} 1^{n} + a_{n - 1} 1^{n - 1} + \dots + a_{1} 1 + a_{0} (mod 6) (N \equiv a_{n} + a_{n - 1} + \dots + a_{1} + a_{0} (mod 6))$ A1

THEN

so $N \equiv 0 (mod 6)$ if and only if $a_{n} + a_{n - 1} + \dots + a_{1} + a_{0} \equiv 0 (mod 6)$ R1

so $6 N$ if and only if $6 (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0})$ AG

METHOD 2

let $N = a_{n} 13^{n} + a_{n - 1} 13^{n - 1} + \dots + a_{1} 13 + a_{0}$ (M1)

$N = (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0}) + (13 - 1) (a_{n} (13^{n - 1} + \dots + 13^{0}) + a_{n - 1} (13^{n - 2} + \dots + 13^{0}) + \dots + a_{1} 13^{0})$ M1A1

Note: Award M1 for attempting to express $N$ in the form $N = (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0}) + (13 - 1) M$ .

as $6 (13 - 1) M$ R1

so $6 N$ if and only if $6 (a_{n} + a_{n - 1} + \dots + a_{1} + a_{0})$ AG

[4 marks]

b.i.

METHOD 1

the sum of the digits is $2 y + 20$ (A1)

uses $2 y + 20 = 6 k$ (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

${(1 y 93 y 25)}_{13} = 1 \times 13^{6} + y \times 13^{5} + 9 \times 13^{4} + 3 \times 13^{3} + y \times 13^{2} + 2 \times 13^{1} + 5 \times 13^{0}$ (A1)

$= 371462 y + 5090480$

attempts to find a valid value of $y$ such that

$371462 y + 5090480 \equiv 0 (mod 6)$ (M1)

$y = 2, 5, 8, 11 (B)$ A1A1

Note: Award A1 for $y = 2, 5, 8$ and A1 for $y = 11 (B)$ .

[4 marks]

b.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

Syllabus sections

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

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Topic 1—Number and algebra