| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Number Theory |
| Type | Divisibility tests and proofs |
| Difficulty | Standard +0.3 This is a straightforward divisibility test requiring students to check divisibility by 2, 3, and 11 (since 66 = 2×3×11). While it requires knowledge of multiple divisibility rules and clear reasoning, it's a direct application of standard techniques with no problem-solving insight needed—slightly easier than average for Further Maths. |
| Spec | 8.02b Divisibility tests: standard tests for 2, 3, 4, 5, 8, 9, 11 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies either 3 or 11 as a prime divisor of 66 and proceeds to apply the divisibility test for this prime number | M1 | Identifies one of the odd prime factors of 66 and proceeds to check divisibility for it |
| Either \(2-0+2-1+0-5+2-0 = 0 = 0\times11\) hence \(n\) is divisible by 11, OR \(2+0+2+1+0+5+2+0 = 12 = 4\times3\) hence \(n\) is divisible by 3 | A1 | Correct method and deduction for either divisibility by 3 or by 11 |
| Both \(2-0+2-1+0-5+2-0 = 0 = 0\times11\) hence \(n\) is divisible by 11, AND \(2+0+2+1+0+5+2+0 = 12 = 4\times3\) hence \(n\) is divisible by 3 | A1 | Correct method and deduction for both divisibility by 3 and by 11 |
| As also \(n\) is even, it is divisible by 2, and hence as divisible by 2, 3 and 11, is divisible by \(2\times3\times11=66\) | A1 | Must be a correct reason for divisibility by 2. Do not accept "last digit is 0" with no reason given. NB No divisibility test for 6 exists |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies either 3 or 11 as a prime divisor of 66 and proceeds to apply the divisibility test for this prime number | **M1** | Identifies one of the odd prime factors of 66 and proceeds to check divisibility for it |
| Either $2-0+2-1+0-5+2-0 = 0 = 0\times11$ hence $n$ is divisible by 11, OR $2+0+2+1+0+5+2+0 = 12 = 4\times3$ hence $n$ is divisible by 3 | **A1** | Correct method and deduction for either divisibility by 3 or by 11 |
| Both $2-0+2-1+0-5+2-0 = 0 = 0\times11$ hence $n$ is divisible by 11, AND $2+0+2+1+0+5+2+0 = 12 = 4\times3$ hence $n$ is divisible by 3 | **A1** | Correct method and deduction for both divisibility by 3 and by 11 |
| As also $n$ is even, it is divisible by 2, and hence as divisible by 2, 3 and 11, is divisible by $2\times3\times11=66$ | **A1** | Must be a correct reason for divisibility by 2. Do not accept "last digit is 0" with no reason given. NB No divisibility test for 6 exists |
---\begin{enumerate}
\item In this question you must show detailed reasoning.
\end{enumerate}
Without performing any division, explain why $n = 20210520$ is divisible by 66
\hfill \mbox{\textit{Edexcel FP2 2021 Q1 [4]}}