Question 2 - A-Level Maths

Edexcel FP2 AS 2019 June — Question 2 7 marks

Exam Board	Edexcel
Module	FP2 AS (Further Pure 2 AS)
Year	2019
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Number Theory
Type	Modular arithmetic properties
Difficulty	Standard +0.8 Part (i) requires understanding that 15 ≡ 3 (mod a) means a divides 12, then finding divisors greater than 3. Part (ii) is a standard modular arithmetic proof using factorization and prime properties. Part (iii) applies divisibility rules (sum of digits for mod 11) creatively. This is a solid Further Maths question requiring multiple modular arithmetic techniques and some problem-solving insight, but the individual components are fairly standard for FP2 level.
Spec	8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 11 8.02e Finite (modular) arithmetic: integers modulo n 8.02l Fermat's little theorem: both forms

(i) Determine all the possible integers $a$, where $a > 3$, such that

$$15 \equiv 3 \bmod a$$ (ii) Show that if $p$ is prime, $x$ is an integer and $x ^ { 2 } \equiv 1 \bmod p$ then either $$x \equiv 1 \bmod p \quad \text { or } \quad x \equiv - 1 \bmod p$$ (iii) A company has $\pounds 13940220$ to share between 11 charities. Without performing any division and showing all your working, decide if it is possible to share this money equally between the 11 charities.

Show mark scheme Show mark scheme source

Question 2:

Part (i)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Any correct value for $a = 4, 6$ or $12$	M1	Understanding of mod notation and finding a correct value
All three correct values $a = 4, 6$ and $12$, no extras	A1	All three correct values, no extras

Part (ii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$x^2 - 1$ is divisible by $p$; OR $x^2-1 \equiv 0 \bmod p$; OR $p \mid (x^2-1)$	B1	See scheme
$\therefore (x-1)(x+1)$ is divisible by $p$ and since $p$ is prime either $(x-1)$ is divisible by $p$ or $(x+1)$ is divisible by $p$; OR $(x-1)(x+1)\equiv 0\bmod p$ and since $p$ is prime either $x-1\equiv 0\bmod p$ or $x+1\equiv 0\bmod p$; OR $p\mid(x-1)(x+1)$ and since $p$ is prime either $p\mid(x-1)$ or $p\mid(x+1)$	M1	Must reference that $p$ is prime
$\therefore\ x\equiv 1\bmod p$ or $x\equiv -1\bmod p$	A1*	Fully correct conclusion

Part (iii)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Selecting and performing a divisibility test for dividing by 11: $1-3+9-4+0-2+2-0=3$; Sum odd $=12$, Sum even $=9$, Difference $=3$ (not 0 or divisible by 11)	M1	Applying divisibility test for dividing by 11 to £13 940 220
3 is not divisible by 11, therefore it is not possible to share this money equally between the 11 charities	A1	Fully correct method with correct sum $(\pm 3)$, concludes not divisible by 11 and interprets in context

## Question 2:

**Part (i)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Any correct value for $a = 4, 6$ or $12$ | M1 | Understanding of mod notation and finding a correct value |
| All three correct values $a = 4, 6$ and $12$, no extras | A1 | All three correct values, no extras |

**Part (ii)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - 1$ is divisible by $p$; OR $x^2-1 \equiv 0 \bmod p$; OR $p \mid (x^2-1)$ | B1 | See scheme |
| $\therefore (x-1)(x+1)$ is divisible by $p$ and **since $p$ is prime** either $(x-1)$ is divisible by $p$ or $(x+1)$ is divisible by $p$; OR $(x-1)(x+1)\equiv 0\bmod p$ and **since $p$ is prime** either $x-1\equiv 0\bmod p$ or $x+1\equiv 0\bmod p$; OR $p\mid(x-1)(x+1)$ and **since $p$ is prime** either $p\mid(x-1)$ or $p\mid(x+1)$ | M1 | Must reference that $p$ is prime |
| $\therefore\ x\equiv 1\bmod p$ or $x\equiv -1\bmod p$ | A1* | Fully correct conclusion |

**Part (iii)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Selecting and performing a divisibility test for dividing by 11: $1-3+9-4+0-2+2-0=3$; Sum odd $=12$, Sum even $=9$, Difference $=3$ (not 0 or divisible by 11) | M1 | Applying divisibility test for dividing by 11 to £13 940 220 |
| 3 is not divisible by 11, therefore it is **not possible** to share this money equally between the 11 charities | A1 | Fully correct method with correct sum $(\pm 3)$, concludes not divisible by 11 and interprets in context |

---

Show LaTeX source

\begin{enumerate}
  \item (i) Determine all the possible integers $a$, where $a > 3$, such that
\end{enumerate}

$$15 \equiv 3 \bmod a$$

(ii) Show that if $p$ is prime, $x$ is an integer and $x ^ { 2 } \equiv 1 \bmod p$ then either

$$x \equiv 1 \bmod p \quad \text { or } \quad x \equiv - 1 \bmod p$$

(iii) A company has $\pounds 13940220$ to share between 11 charities.

Without performing any division and showing all your working, decide if it is possible to share this money equally between the 11 charities.

\hfill \mbox{\textit{Edexcel FP2 AS 2019 Q2 [7]}}

This paper (5 questions)

View full paper

Q1 5 Q2 7 Q3 10 Q4 7 Q5 11