| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Proofs |
| Type | Proof by exhaustion |
| Difficulty | Moderate -0.8 This question requires minimal mathematical sophistication: part (a) needs finding any counterexample (e.g., 13²=169, 1+3=4≠9), part (b) is trivial pattern recognition (9²→last digit 1), and part (c) is a routine proof by exhaustion checking 0²,1²,...,9² mod 10. Despite being labeled 'proof by exhaustion,' it's essentially checking 10 cases with single-digit arithmetic—well below average A-level difficulty. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.01c Disproof by counter example |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(12^2 = 144\), digit sum \(= 1+2 = 3\) | M1 | Squares a number with two or more digits and adds its digits; must be explicit |
| \(3 \neq 4\) | R1 | Completes argument to show Asif's method is incorrect; must compare sum of digits with last digit of square number |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1\) | B1 | Obtains 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0^2=0,\ 1^2=1,\ 2^2=4,\ 3^2=9,\ 4^2=16,\ 5^2=25,\ 6^2=36,\ 7^2=49,\ 8^2=64,\ 9^2=81\) | M1 | Lists at least four single digits and their squares; or explains why odd digits do not need to be considered |
| Therefore there can be no square number which has a last digit of 8 | R1 | Completes rigorous argument; OE; CSO |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $12^2 = 144$, digit sum $= 1+2 = 3$ | M1 | Squares a number with two or more digits and adds its digits; must be explicit |
| $3 \neq 4$ | R1 | Completes argument to show Asif's method is incorrect; must compare sum of digits with last digit of square number |
---
## Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1$ | B1 | Obtains 1 |
---
## Question 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0^2=0,\ 1^2=1,\ 2^2=4,\ 3^2=9,\ 4^2=16,\ 5^2=25,\ 6^2=36,\ 7^2=49,\ 8^2=64,\ 9^2=81$ | M1 | Lists at least four single digits and their squares; or explains why odd digits do not need to be considered |
| Therefore there can be no square number which has a last digit of 8 | R1 | Completes rigorous argument; OE; CSO |
---6
\begin{enumerate}[label=(\alph*)]
\item Asif notices that $24 ^ { 2 } = 576$ and $2 + 4 = 6$ gives the last digit of 576
He checks two more examples:
$$\begin{array} { l c }
27 ^ { 2 } = 729 & 29 ^ { 2 } = 841 \\
2 + 7 = 9 & 2 + 9 = 11 \\
\text { Last digit } 9 & \text { Last digit } 1
\end{array}$$
Asif concludes that he can find the last digit of any square number greater than 100 by adding the digits of the number being squared.
Give a counter example to show that Asif's conclusion is not correct.
6
\item Claire tells Asif that he should look only at the last digit of the number being squared.
$$\begin{array} { c c }
27 ^ { 2 } = 729 & 24 ^ { 2 } = 576 \\
7 ^ { 2 } = 49 & 4 ^ { 2 } = 16 \\
\text { Last digit } 9 & \text { Last digit } 6
\end{array}$$
Using Claire's method determine the last digit of $23456789 { } ^ { 2 }$\\[0pt]
[1 mark]
6
\item Given Claire's method is correct, use proof by exhaustion to show that no square number has a last digit of 8
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q6 [5]}}