AQA Further Paper 3 Discrete 2022 June — Question 9 6 marks

Exam BoardAQA
ModuleFurther Paper 3 Discrete (Further Paper 3 Discrete)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGroups
TypeProve group-theoretic identities
DifficultyStandard +0.8 This is a Further Maths discrete mathematics question on binary operations requiring understanding of commutativity, associativity, and identity elements. Part (a)(i) is routine, but (a)(ii) requires careful algebraic manipulation with modular arithmetic, and part (b) involves solving a quadratic congruence condition. The multi-step reasoning and abstract algebraic concepts place it moderately above average difficulty, though it follows standard patterns for Further Maths group theory questions.
Spec8.02e Finite (modular) arithmetic: integers modulo n8.03a Binary operations: and their properties on given sets8.03b Cayley tables: construct for finite sets under binary operation

The binary operation \(\oplus\) acts on the positive integers \(x\) and \(y\) such that $$x \oplus y = x + y + 8 \pmod{k^2 - 16k + 74}$$ where \(k\) is a positive integer.
    1. Show that \(\oplus\) is commutative. [1 mark]
    2. Determine whether or not \(\oplus\) is associative. Fully justify your answer. [2 marks]
  1. Find the values of \(k\) for which 3 is an identity element for the set of positive integers under \(\oplus\) [3 marks]

Question 9:

AnswerMarks
9(a)(i)Deduces the operation is
commutative by considering
both x y and y  x and
shows that they are the same
Condone ( m o d k 2 − 1 6 k + 7 4 )
AnswerMarks Guidance
missing2.2a B1
y  x = y + x + 8 ( m o d k 2 − 1 6 k + 7 4 )
= x + y + 8 ( m o d k 2 − 1 6 k + 7 4 )
AnswerMarks Guidance
Total1
QMarking instructions AO

AnswerMarks
9(a)(ii)Sets up test for associativity by
considering combinations of the
form ( x  y )  z and x  ( y  z )
and simplifies one combination
to obtain x + y + z + 1 6
Condone ( m o d k 2 − 1 6 k + 7 4 )
AnswerMarks Guidance
missing1.1a M1
= ( ) 8 x + y + + z + 8 ( m o d k 2 − 1 6 k + 7 4 )
 ( ) x  y  z
= 1 6 x + y + z + ( m o 2 d 1 k − 6 k + 7 4 )
x ( )  y  z
= ( 8 x + y + z + ) + 8 ( m o d k 2 − 1 6 k + 7 4 )
 ( ) x  y  z
= 1 6 x + y + z + ( m o 2 d 1 k − 6 k + 7 4 )
 ( ) x  y  z = ( x  ) y  z
Hence, the binary operation is
associative.
Completes a reasoned
argument and concludes that
the binary operation is
associative.
Condone ( m o d k 2 − 1 6 k + 7 4 )
AnswerMarks Guidance
missing2.1 R1
Total2
QMarking instructions AO

AnswerMarks
9(b)Translates problem by applying
binary operation between 3 and
another positive integer
Condone ( m o d k 2 − 1 6 k + 7 4 )
AnswerMarks Guidance
missing3.1a M1
3  x = x + 1 1 ( m o d k 2 − 1 6 k + 7 4 )
3x= x ( mod k2−16k+74 )
 k2−16k+74=11
k2−16k+63=0
k =7,9
Uses the condition for 3 to be an
identity element to deduce that
AnswerMarks Guidance
the addition must be modulo 112.2a A1
Solves the quadratic equation to
find the two correct values for k
AnswerMarks Guidance
and no others3.2a A1
Total3
Question total6
QMarking instructions AO
Question 9:
--- 9(a)(i) ---
9(a)(i) | Deduces the operation is
commutative by considering
both x y and y  x and
shows that they are the same
Condone ( m o d k 2 − 1 6 k + 7 4 )
missing | 2.2a | B1 | x  y = x + y + 8 ( m o d k 2 − 1 6 k + 7 4 )
y  x = y + x + 8 ( m o d k 2 − 1 6 k + 7 4 )
= x + y + 8 ( m o d k 2 − 1 6 k + 7 4 )
Total | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 9(a)(ii) ---
9(a)(ii) | Sets up test for associativity by
considering combinations of the
form ( x  y )  z and x  ( y  z )
and simplifies one combination
to obtain x + y + z + 1 6
Condone ( m o d k 2 − 1 6 k + 7 4 )
missing | 1.1a | M1 | ( ) x  y  z
= ( ) 8 x + y + + z + 8 ( m o d k 2 − 1 6 k + 7 4 )
 ( ) x  y  z
= 1 6 x + y + z + ( m o 2 d 1 k − 6 k + 7 4 )
x ( )  y  z
= ( 8 x + y + z + ) + 8 ( m o d k 2 − 1 6 k + 7 4 )
 ( ) x  y  z
= 1 6 x + y + z + ( m o 2 d 1 k − 6 k + 7 4 )
 ( ) x  y  z = ( x  ) y  z
Hence, the binary operation is
associative.
Completes a reasoned
argument and concludes that
the binary operation is
associative.
Condone ( m o d k 2 − 1 6 k + 7 4 )
missing | 2.1 | R1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Translates problem by applying
binary operation between 3 and
another positive integer
Condone ( m o d k 2 − 1 6 k + 7 4 )
missing | 3.1a | M1 | 3  x = 3 + x + 8 ( m o d k 2 − 1 6 k + 7 4 )
3  x = x + 1 1 ( m o d k 2 − 1 6 k + 7 4 )
3x= x ( mod k2−16k+74 )
 k2−16k+74=11
k2−16k+63=0
k =7,9
Uses the condition for 3 to be an
identity element to deduce that
the addition must be modulo 11 | 2.2a | A1
Solves the quadratic equation to
find the two correct values for k
and no others | 3.2a | A1
Total | 3
Question total | 6
Q | Marking instructions | AO | Marks | Typical solution
The binary operation $\oplus$ acts on the positive integers $x$ and $y$ such that

$$x \oplus y = x + y + 8 \pmod{k^2 - 16k + 74}$$

where $k$ is a positive integer.

\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\oplus$ is commutative.
[1 mark]

\item Determine whether or not $\oplus$ is associative.

Fully justify your answer.
[2 marks]
\end{enumerate}

\item Find the values of $k$ for which 3 is an identity element for the set of positive integers under $\oplus$
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Further Paper 3 Discrete 2022 Q9 [6]}}