June 2022 Paper 2 Q6
6
(a) Asif notices that \(24^2 = 576\) and \(2 + 4 = 6\) gives the last digit of 576
He checks two more examples:
| \(27^2 = 729\) | \(29^2 = 841\) |
| \(2 + 7 = 9\) | \(2 + 9 = 11\) |
| Last digit 9 | Last digit 1 |
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. [2 marks]
(b) Claire tells Asif that he should look only at the last digit of the number being squared.
| \(27^2 = 729\) | \(24^2 = 576\) |
| \(7^2 = 49\) | \(4^2 = 16\) |
| Last digit 9 | Last digit 6 |
Using Claire’s method determine the last digit of \(23456789^2\) [1 mark]
(c) Given Claire’s method is correct, use proof by exhaustion to show that no square number has a last digit of 8 [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Squares a number with two or more digits and adds its digits Must be explicit | M1 | 1.1a |
| Completes argument to show that Asif’s method is incorrect Must compare sum of digits with last digit of square number | R1 | 2.3 |
| (2) |
Typical solution
\[12^2 = 144\]\[1 + 2 = 3\]\[3 \ne 4\]| Scheme | Marks | AO |
|---|---|---|
| Obtains 1 | B1 | 1.1b |
| (1) |
Typical solution
1
| Scheme | Marks | AO |
|---|---|---|
| Lists at least four single digits and their squares Or Explains why odd digits do not need to be considered | M1 | 1.1a |
| Completes rigorous argument to prove that no square number has a last digit of 8 OE CSO | R1 | 2.1 |
| (2) | ||
| (5 marks) |
Typical solution
| \(0^2 = 0\) | \(4^2 = 16\) | \(8^2 = 64\) |
| \(1^2 = 1\) | \(5^2 = 25\) | \(9^2 = 81\) |
| \(2^2 = 4\) | \(6^2 = 36\) | |
| \(3^2 = 9\) | \(7^2 = 49\) |
Therefore, there can be no square number which has a last digit of 8