| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Subgroups and cosets |
| Difficulty | Challenging +1.8 This question requires knowledge of Fermat's Little Theorem, Lagrange's theorem (subgroup order divides group order), and modular arithmetic to determine 46^46 + 47^47 (mod 11) and (mod 21). While the concepts are A-level appropriate, applying FLT in this context requires mathematical maturity and multi-step reasoning beyond standard textbook exercises, making it significantly harder than average but not at the extreme difficulty level of proof-heavy AEA questions. |
| Spec | 8.02l Fermat's little theorem: both forms8.03k Lagrange's theorem: order of subgroup divides order of group |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Order of subgroup must divide order of group (Lagrange's Theorem), need to check if 11 (and/or 21) divides \(46^{46}+47^{47}\); by FLT \(a^{11-1}=a^{10}\equiv1\pmod{11}\) | M1 | Attempt to apply Fermat's Little Theorem at least once with \(p=11\), \(p=7\) or \(p=3\) |
| \(46^{46}+47^{47}\equiv2^{4\times10+6}+3^{4\times10+7}\equiv2^6+3^7\equiv64+(3^3)^2\times3 \equiv9+5^2\times3\equiv84\equiv7\pmod{11}\) | M1 | Applies FLT and congruence arithmetic fully to find residue of \(46^{46}+47^{47}\) modulo 11 |
| Hence 11 is not a divisor of \(46^{46}+47^{47}\), so not a possible order for a subgroup | A1 | \(46^{46}+47^{47}\equiv7\pmod{11}\), deduces not a possible order |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(21=7\times3\) so need to check for factors of 7 and 3, using \(a^2\equiv1\pmod3\) and \(a^6\equiv1\pmod7\) | M1 | Applies checks for both 7 and 3 as divisors via similar strategy |
| \(46^{46}+47^{47}\equiv1^{46}+2^{47}\equiv1+2^{2\times23+1}\equiv1+2^1\equiv3\equiv0\pmod3\) | M1 | Applies FLT with \(p=3\) to find smaller residue modulo 3 |
| \(46^{46}+47^{47}\equiv4^{46}+(-2)^{47}\equiv4^{6\times7+4}+(-2)^{6\times7+5}\equiv4^4+(-2)^5 \equiv16^2-32\equiv9^2-4\equiv81-4\equiv77\equiv0\pmod7\) | M1 | Applies FLT with \(p=7\) to find smaller residue modulo 7 |
| As \(46^{46}+47^{47}\) divisible by both 3 and 7, it is divisible by 21 and hence this is a possible order for a subgroup | A1 | Shows congruent to 0 mod 3 and mod 7, deduces 21 divides \(46^{46}+47^{47}\), hence possible order |
| (4) |
# Question 4:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| Order of subgroup must divide order of group (Lagrange's Theorem), need to check if 11 (and/or 21) divides $46^{46}+47^{47}$; by FLT $a^{11-1}=a^{10}\equiv1\pmod{11}$ | **M1** | Attempt to apply Fermat's Little Theorem at least once with $p=11$, $p=7$ or $p=3$ |
| $46^{46}+47^{47}\equiv2^{4\times10+6}+3^{4\times10+7}\equiv2^6+3^7\equiv64+(3^3)^2\times3 \equiv9+5^2\times3\equiv84\equiv7\pmod{11}$ | **M1** | Applies FLT and congruence arithmetic fully to find residue of $46^{46}+47^{47}$ modulo 11 |
| Hence 11 is not a divisor of $46^{46}+47^{47}$, so not a possible order for a subgroup | **A1** | $46^{46}+47^{47}\equiv7\pmod{11}$, deduces not a possible order |
| | **(3)** | |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $21=7\times3$ so need to check for factors of 7 and 3, using $a^2\equiv1\pmod3$ and $a^6\equiv1\pmod7$ | **M1** | Applies checks for both 7 and 3 as divisors via similar strategy |
| $46^{46}+47^{47}\equiv1^{46}+2^{47}\equiv1+2^{2\times23+1}\equiv1+2^1\equiv3\equiv0\pmod3$ | **M1** | Applies FLT with $p=3$ to find smaller residue modulo 3 |
| $46^{46}+47^{47}\equiv4^{46}+(-2)^{47}\equiv4^{6\times7+4}+(-2)^{6\times7+5}\equiv4^4+(-2)^5 \equiv16^2-32\equiv9^2-4\equiv81-4\equiv77\equiv0\pmod7$ | **M1** | Applies FLT with $p=7$ to find smaller residue modulo 7 |
| As $46^{46}+47^{47}$ divisible by both 3 and 7, it is divisible by 21 and hence this is a possible order for a subgroup | **A1** | Shows congruent to 0 mod 3 and mod 7, deduces 21 divides $46^{46}+47^{47}$, hence possible order |
| | **(4)** | |
---\begin{enumerate}
\item Let $G$ be a group of order $46 ^ { 46 } + 47 ^ { 47 }$
\end{enumerate}
Using Fermat's Little Theorem and explaining your reasoning, determine which of the following are possible orders for a subgroup of $G$\\
(i) 11\\
(ii) 21
\hfill \mbox{\textit{Edexcel FP2 2021 Q4 [7]}}