Question 5 - A-Level Maths

OCR Further Additional Pure AS 2023 June — Question 5 11 marks

Exam Board	OCR
Module	Further Additional Pure AS (Further Additional Pure AS)
Year	2023
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Number Theory
Type	Base conversion
Difficulty	Standard +0.8 This is a Further Maths question combining base conversion, linear congruences, and divisibility proofs. Part (a) is routine calculation, part (b) requires modular arithmetic with multiplicative inverses, and part (c) demands algebraic proof and iterative application. The multi-step nature and proof requirement elevate it above standard A-level, but the techniques are well-established for Further Maths students.
Spec	8.02a Number bases: conversion and arithmetic in base n 8.02f Single linear congruences: solve ax = b (mod n)

Express as a decimal (base-10) number the base-23 number \(7119 _ { 23 }\).
Solve the linear congruence \(7 n + 11 \equiv 9 ( \bmod 23 )\).
Let \(N = 10 a + b\) and \(M = a + 7 b\), where \(a\) and \(b\) are integers and \(0 \leqslant b \leqslant 9\).
1. By considering \(3 N - 7 M\), prove that \(23 \mid N\) if and only if \(23 \mid M\).
2. Use a procedure based on this result to show that \(N = 711965\) is a multiple of 23 .

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5	(a)	7 1 1 9 = 9 + (1  23) + (1  232) + (7  233)

2 3

= 85 730

Answer	Marks
10	M1

A1

Answer	Marks	Guidance
[2]	1.1
1.1	Clear use of base-23 column-values
5	(b)	7n + 11  9 (mod 23)  7n + 11  32 (mod 23)

OR 7n  −2 (mod 23)

 7n  21 (mod 23)

 n  3 (mod 23) or n = 23k + 3 (k  ℤ)

Answer	Marks
hcf(7, 23) = 1 for division to be valid	M1

A1

B1

Answer	Marks
[3]	1.1

1.1

Answer	Marks
2.4	Use of modular arithmetic to gain a multiple of 7 on

the RHS

Any correct, complete statement

Accept n  −20 (mod 23)

Statement that hcf(7, 23) = 1 for division to be valid

or equivalent

Answer	Marks	Guidance
5	(c)	i

If N = 23k, then 7M = 23(3k – a + 2b)

If M = 23k, then 3N = 23(7k + a – 2b)

Both correctly shown multiples of 23 with explanation that

Answer	Marks
hcf(7, 23) = 1 and hcf(3, 23) = 1	B1

M1

A1

Answer	Marks
[4]	1.1

2.1

Answer	Marks
2.4	Must be written explicitly as a multiple of 23, here or

later on

Proof attempted in one direction

(“proof” includes attempt to obtain a multiple of 23)

Proof attempted in other direction

(“proof” includes attempt to obtain a multiple of 23)

At least one justification must be noted

[2]

Question 5:
5 | (a) | 7 1 1 9 = 9 + (1  23) + (1  232) + (7  233)
2 3
= 85 730
10 | M1
A1
[2] | 1.1
1.1 | Clear use of base-23 column-values
5 | (b) | 7n + 11  9 (mod 23)  7n + 11  32 (mod 23)
OR 7n  −2 (mod 23)
 7n  21 (mod 23)
 n  3 (mod 23) or n = 23k + 3 (k  ℤ)
hcf(7, 23) = 1 for division to be valid | M1
A1
B1
[3] | 1.1
1.1
2.4 | Use of modular arithmetic to gain a multiple of 7 on
the RHS
Any correct, complete statement
Accept n  −20 (mod 23)
Statement that hcf(7, 23) = 1 for division to be valid
or equivalent
5 | (c) | i | 3N – 7M = 30a + 3b – 7a – 49b = 23(a – 2b)
If N = 23k, then 7M = 23(3k – a + 2b)
If M = 23k, then 3N = 23(7k + a – 2b)
Both correctly shown multiples of 23 with explanation that
hcf(7, 23) = 1 and hcf(3, 23) = 1 | B1
M1
M1
A1
[4] | 1.1
2.1
2.1
2.4 | Must be written explicitly as a multiple of 23, here or
later on
Proof attempted in one direction
(“proof” includes attempt to obtain a multiple of 23)
Proof attempted in other direction
(“proof” includes attempt to obtain a multiple of 23)
At least one justification must be noted
[2]

Show LaTeX source

5
\begin{enumerate}[label=(\alph*)]
\item Express as a decimal (base-10) number the base-23 number $7119 _ { 23 }$.
\item Solve the linear congruence $7 n + 11 \equiv 9 ( \bmod 23 )$.
\item Let $N = 10 a + b$ and $M = a + 7 b$, where $a$ and $b$ are integers and $0 \leqslant b \leqslant 9$.
\begin{enumerate}[label=(\roman*)]
\item By considering $3 N - 7 M$, prove that $23 \mid N$ if and only if $23 \mid M$.
\item Use a procedure based on this result to show that $N = 711965$ is a multiple of 23 .
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2023 Q5 [11]}}

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