| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | March |
| Marks | 4 |
| Topic | Number Theory |
| Type | Divisibility tests and proofs |
| Difficulty | Moderate -0.8 This question requires only direct application of standard divisibility tests (sum of digits for 9, alternating sum for 11, last three digits for 8) with no problem-solving or insight needed. While it involves large numbers requiring careful arithmetic, the tests are mechanical procedures taught explicitly in the syllabus, making this easier than average despite being Further Maths content. |
| Spec | 8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 118.02c Divisibility by primes: algorithmic tests for primes less than 50 |
| Answer | Marks | Guidance |
|---|---|---|
| B1, M1, A1 | 1.1a, 2.1, 1.1 | Or Digit-sum of \(N\) is 54 and \(5 + 4 = 9\) |
| Or \(588 \div 2 = 294, 294 \div 2 = 147\) which is odd, hence \(8 \nmid N\) | ||
| B1 | 1.1a | |
| [4] |
**Answer:** Digit-sum of $N$ is 54 and since $9 \mid 54, 9 \mid N$. Alternate digit-sums are … odds: 38, evens: 16, 38 − 16 = 22. Since $11 \mid$ the difference (or 22), $11 \mid N$
$8 \mid N$ if and only if $8 \mid$ (the last-three-digit part of $N$) and $\frac{588}{8} = 73.5$
hence $8 \nmid N$
| **B1, M1, A1** | **1.1a, 2.1, 1.1** | Or Digit-sum of $N$ is 54 and $5 + 4 = 9$
Or $588 \div 2 = 294, 294 \div 2 = 147$ which is odd, hence $8 \nmid N$ |
| **B1** | **1.1a** | |
| **[4]** | | |
---1 Use standard divisibility tests to show that the number
$$N = 91039173588$$
\begin{itemize}
\item is divisible by 9
\item is divisible by 11
\item is not divisible by 8 .
\end{itemize}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q1 [4]}}