| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Divisibility tests and proofs |
| Difficulty | Challenging +1.2 This is a Further Maths question on divisibility proofs requiring algebraic manipulation to show M+2N=102a-17b is divisible by 17, then applying the algorithm iteratively. While it requires understanding of modular arithmetic and proof structure, the algebraic steps are straightforward and the algorithm application is mechanical. It's moderately harder than average A-level questions due to the proof component and Further Maths context, but not exceptionally challenging. |
| Spec | 8.02d Division algorithm: a = bq + r uniquely |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | M + 2N = 102a – 17b = 17(6a – b) |
| If 17 | N then 17 | M = 17(…) – 2N |
| If 17 | M then 17 | 2N = 17(…) – M 17 |
| since hcf(2, 17) = 1 | B1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 2.1 | The factor of 17 must be noted (here |
| Answer | Marks |
|---|---|
| For valid argument | Condone lack of use of |
| Answer | Marks |
|---|---|
| (ii) | 2 058 376 813 901 20 583 768 139 – 9 |
| Answer | Marks |
|---|---|
| and the original number M is a multiple of 17 | M1 A1 |
| Answer | Marks |
|---|---|
| A1 | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | First step clearly and correctly |
Question 5:
5 | (i) | M + 2N = 102a – 17b = 17(6a – b)
If 17 | N then 17 | M = 17(…) – 2N
If 17 | M then 17 | 2N = 17(…) – M 17 | N
since hcf(2, 17) = 1 | B1
M1
A1
A1 | 1.1
3.1a
2.4
2.1 | The factor of 17 must be noted (here
or later)
For valid argument
For valid argument | Condone lack of use of
Euclid’s Lemma to justify
result
[4]
(ii) | 2 058 376 813 901 20 583 768 139 – 9
= 20 583 768 130
20 583 768 130 205 837 681 – 270
= 205 837 411
205 837 411 2 058 374 – 99 = 2 058 275
2 058 275 20 582 – 675 = 19 907
19 907 199 – 63 = 136
136 = 8 17
and the original number M is a multiple of 17 | M1 A1
M1
A1 | 1.1a
1.1
1.1
1.1 | First step clearly and correctly
demonstrated
Process repeated iteratively to a
suitable stopping-point
All correctly demonstrated
Final statement of result not required
[4]5 For integers $a$ and $b$, with $a \geqslant 0$ and $0 \leqslant b \leqslant 99$, the numbers $M$ and $N$ are such that
$$M = 100 a + b \text { and } N = a - 9 b .$$
(i) By considering the number $M + 2 N$, show that $17 \mid M$ if and only if $17 \mid N$.\\
(ii) Demonstrate step-by-step how an algorithm based on the result of part (i) can be used to show that 2058376813901 is a multiple of 17 .
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q5 [8]}}