A2 October 2021 Q1
1.
In this question you must show detailed reasoning.
Without performing any division, explain why \(n = 20\,210\,520\) is divisible by 66 (4)
| Scheme | Marks | AO |
|---|---|---|
| Identifies either 3 or 11 as a prime divisor of 66 and proceeds to apply the divisibility test for this prime number. | M1 | 3.1a |
| Either \(2 - 0 + 2 - 1 + 0 - 5 + 2 - 0 = 0 = 0 \times 11\) hence \(n\) is divisible by 11 Or \(2 + 0 + 2 + 1 + 0 + 5 + 2 + 0 = 12 = 4 \times 3\) hence \(n\) is divisible by 3 | A1 | 2.2a |
| Both \(2 - 0 + 2 - 1 + 0 - 5 + 2 - 0 = 0 = 0 \times 11\) hence \(n\) is divisible by 11 And \(2 + 0 + 2 + 1 + 0 + 5 + 2 + 0 = 12 = 4 \times 3\) hence \(n\) is divisible by 3 | A1 | 2.2a |
| As also \(n\) is even, it is divisible by 2, and hence as divisible by 2, 3 and 11, is divisible by \(2 \times 3 \times 11 = 66\) | A1 | 2.4 |
| (4) | ||
| (4 marks) |
Notes
M1: Identifies one of the odd prime factors of 66 and proceeds to check divisibility for it.
A1: Correct method and deduction for either divisibility by 3 or by 11
A1: Correct method and deduction for both divisibility by 3 and by 11
A1: Notes also divisibility by 2 and explains why divisibility by 66 follows. The explanation may have been given in a preamble “66 = 2×3×11 so divisible by 66 if divisible by 2, 3 and 11”
The must be a correct reason for divisibility by 2, ie “it is even” or “last digit is even”. Do not accept “last digit is 0” with no reason given.
NB There is no divisibility test for 6, so attempting such will result in the loss of the last two A marks. (E.g. sum of digits being a multiple of 6 is an incorrect test, as for instance 15 satisfies 1+5 = 6 but is not a multiple of 6.)