Challenging +1.2 This is a Further Maths question requiring proof that gcd(4n+1, 6n+1)=1 using the Euclidean algorithm or contradiction. While it requires understanding of coprimality and algebraic manipulation (finding a linear combination equaling 1), the technique is standard and the algebra straightforward: 3(4n+1) - 2(6n+1) = 1. More challenging than routine A-level due to proof requirement and Further Maths context, but not exceptionally difficult.
a = 3, b = 2 so that n is eliminated, “1” obtained
Answer
Marks
Guidance
so if h = hcf(4n + 1, 6n + 1) then h
1 h = 1
M1
A1
Answer
Marks
A1
1.2
1.1
Answer
Marks
2.2a
A linear combination of the 2 terms
Given Answer properly deduced
a , b must be integers
[3]
Alt. Let h = hcf(4n + 1, 6n + 1).
Answer
Marks
Then h
{6(4n + 1) – 4(6n + 1)} = 2 h = 1 or 2
But (4n + 1), (6n + 1) are odd, so h 2 h = 1
M 1 A 1
E 1
T w o parts required for the M
Alt. Euclidean Algorithm: 6n+1 = 1(4n+1) + 2n
4n + 1 = 2(2n) + 1
2n = 2n(1) + 0
Answer
Marks
h = final non-zero remainder = 1
M1
A1
Answer
Marks
A1
Correct to at least here
Given Answer properly deduced
Question 3:
3 | a(4n + 1) – b(6n + 1) attempted
a = 3, b = 2 so that n is eliminated, “1” obtained
so if h = hcf(4n + 1, 6n + 1) then h | 1 h = 1 | M1
A1
A1 | 1.2
1.1
2.2a | A linear combination of the 2 terms
Given Answer properly deduced | a , b must be integers
[3]
Alt. Let h = hcf(4n + 1, 6n + 1).
Then h | {6(4n + 1) – 4(6n + 1)} = 2 h = 1 or 2
But (4n + 1), (6n + 1) are odd, so h 2 h = 1 | M 1 A 1
E 1 | T w o parts required for the M
Alt. Euclidean Algorithm: 6n+1 = 1(4n+1) + 2n
4n + 1 = 2(2n) + 1
2n = 2n(1) + 0
h = final non-zero remainder = 1 | M1
A1
A1 | Correct to at least here
Given Answer properly deduced
3 Given that $n$ is a positive integer, show that the numbers ( $4 n + 1$ ) and ( $6 n + 1$ ) are co-prime.
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q3 [3]}}