Moderate -0.3 Part (i) requires checking only four cases (p=2,3,5,7) or factoring p³+p=p(p²+1) and using modular arithmetic—straightforward for A-level. Part (ii) is routine algebraic expansion showing (n+1)³-n³=3n²+3n+1≡1(mod 3). Both parts are direct applications of basic techniques with no novel insight required, making this slightly easier than average.
3. (i) Prove that for all single digit prime numbers, \(p\),
$$p ^ { 3 } + p \text { is a multiple of } 10$$
(ii) Show, using algebra, that for \(n \in \mathbb { N }\)
$$( n + 1 ) ^ { 3 } - n ^ { 3 } \text { is not a multiple of } 3$$
Expands to four-term cubic and cancels \(n^3\) terms; correct quadratic in \(n\)
\(= 3(n^2 + n) + 1\) which is one more than a multiple of 3, so is not divisible by 3 for any \(n \in \mathbb{N}\)
A1
Correct explanation and conclusion; accept factoring out 3 to achieve \(3\times(n^2+n)+1\), or explaining each term other than 1 is divisible by 3
(3)
## Question 3:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Checks at least three of: $p=2: 2^3+2=10$; $p=3: 3^3+3=30$; $p=5: 5^3+5=130$; $p=7: 7^3+7=350$ | **M1** | At least three of the four single digit primes checked; attaining a multiple of 10 is enough |
| Each case gives a multiple of 10; as 2,3,5 and 7 are the only single digit primes, result proved for all single digit primes | **A1** | All four cases correct with minimal conclusion; ignore checks on non-primes; award A0 if $p=1$ leads to conclusion result is not true |
| | **(2)** | |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3n^2 + 3n + 1$ | **M1A1** | Expands to four-term cubic and cancels $n^3$ terms; correct quadratic in $n$ |
| $= 3(n^2 + n) + 1$ which is one more than a multiple of 3, so is not divisible by 3 for any $n \in \mathbb{N}$ | **A1** | Correct explanation and conclusion; accept factoring out 3 to achieve $3\times(n^2+n)+1$, or explaining each term other than 1 is divisible by 3 |
| | **(3)** | |
3. (i) Prove that for all single digit prime numbers, $p$,
$$p ^ { 3 } + p \text { is a multiple of } 10$$
(ii) Show, using algebra, that for $n \in \mathbb { N }$
$$( n + 1 ) ^ { 3 } - n ^ { 3 } \text { is not a multiple of } 3$$
\hfill \mbox{\textit{Edexcel P2 2021 Q3 [5]}}