| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Modular arithmetic properties |
| Difficulty | Challenging +1.2 This is a Further Maths modular arithmetic question requiring Fermat's Little Theorem (or pattern recognition) for part (a) and systematic application of divisibility rules for part (b). While it requires knowledge beyond standard A-level, the techniques are straightforward applications of taught methods without requiring novel insight or complex multi-step reasoning. |
| Spec | 8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 118.02e Finite (modular) arithmetic: integers modulo n |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | ( )n ( )n |
| Answer | Marks |
|---|---|
| = 11×5n ≡ 0 (mod 11), as required | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | Appropriate indices work attempted |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | (i) | 1033 + 1 = a “1” followed by 32 “0”s and a final “1” |
| Answer | Marks | Guidance |
|---|---|---|
| and 11 | 0 ⇒ divisibility by 11, as required | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 1.1 | It is only necessary to note that there is an even number of “0”s |
| Answer | Marks |
|---|---|
| (ii) | Factoring out the first 11 |
| Answer | Marks |
|---|---|
| factor of 11, i.e. 121 | M1 |
| Answer | Marks |
|---|---|
| A1 | 3.1a |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 1033 + 1 = (1011 + 1)(1022 – 1011 + 1) | M1 | NB (1011 + 1)/121 = 826 446 281 ⇒ 121 |
| Showing 11 | 1011 + 1, as in (b) (i) | A1 |
| (1011 + 1)/11 = 9 090 909 091 = 11 × 826 446 281 | M1 A1 | Or via a second use of the 11-divisibility test: |
Question 7:
7 | (a) | ( )n ( )n
f(n) = 24 ×23 + 33 ×3
8×16n +3×27n
=
≡ 8×5n +3×5n (mod 11)
= 11×5n ≡ 0 (mod 11), as required | M1
A1
M1
A1
[4] | 3.1a
1.1
2.1
2.2a | Appropriate indices work attempted
Fully correct
Reducing all terms to mod 11
AG fully shown
(b) | (i) | 1033 + 1 = a “1” followed by 32 “0”s and a final “1”
Then “(Σodds) – (Σevens)” = 1 – 1 = 0
and 11 | 0 ⇒ divisibility by 11, as required | B1
B1
[2] | 1.1a
1.1 | It is only necessary to note that there is an even number of “0”s
between the first and last “1”
Proper explanation of the divisibility test
(ii) | Factoring out the first 11
(1033 + 1)/11 = 15 “90”s followed by a final “91”
Then “(Σodds) – (Σevens)” = 16 × 9 – 1 = 143
which is a multiple of 11 ⇒ divisibility by another
factor of 11, i.e. 121 | M1
A1
M1
A1 | 3.1a
1.1
1.1
1.1 | BC
Conclusion with explanation
Alternative method
1033 + 1 = (1011 + 1)(1022 – 1011 + 1) | M1 | NB (1011 + 1)/121 = 826 446 281 ⇒ 121 | 1033 + 1
Showing 11 | 1011 + 1, as in (b) (i) | A1 | Must use the 11-divisibility test at least once
(1011 + 1)/11 = 9 090 909 091 = 11 × 826 446 281 | M1 A1 | Or via a second use of the 11-divisibility test:
“(Σodds) – (Σevens)” = 5 × 9 – 1 = 44
[4]7
\begin{enumerate}[label=(\alph*)]
\item Let $f ( n ) = 2 ^ { 4 n + 3 } + 3 ^ { 3 n + 1 }$.
Use arithmetic modulo 11 to prove that $\mathrm { f } ( n ) \equiv 0 ( \bmod 11 )$ for all integers $n \geqslant 0$.
\item Use the standard test for divisibility by 11 to prove the following statements.
\begin{enumerate}[label=(\roman*)]
\item $10 ^ { 33 } + 1$ is divisible by 11
\item $10 ^ { 33 } + 1$ is divisible by 121
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2021 Q7 [10]}}