OCR Further Additional Pure AS 2017 December — Question 1 3 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2017
SessionDecember
Marks3
TopicNumber Theory
TypeLinear congruences (single)
DifficultyModerate -0.5 This is a straightforward linear congruence requiring students to find gcd(12,99)=3, verify divisibility, simplify to 4x≡1(mod 33), then find the multiplicative inverse of 4 modulo 33. While it's a Further Maths topic, it's a standard textbook exercise with a clear algorithmic approach requiring no novel insight—slightly easier than average due to its routine nature and small numbers.
Spec8.02f Single linear congruences: solve ax = b (mod n)
1 Solve \(12 x \equiv 3 ( \bmod 99 )\).
AnswerMarks Guidance
12x ≡ 3 (mod 99) ⟹ 4x ≡ 1 (mod 33) ⟹ 4x = 1, 34, 67, 100, ... ⟹ x ≡ 25 (mod 33)B1 M1 A1 Considering equivalents mod 33; Or in the form \(x = 33n + 25\) (\(n \in \mathbb{Z}\)); Or 25, 58, 91 (mod 99); For \(4x \equiv 1\) (mod 99) B0 Then \(4x = 1, 34, 67, 100, ...\) M1 (or via two steps) ⟹ \(x \equiv 25\) (mod 99) or \(x = 99n + 25\) (\(n \in \mathbb{Z}\)) A1
or
AnswerMarks Guidance
99 = 8×12 + 3; So -8 is a solution and hcf(99,12)=3 ⟹ x ≡ -8 (mod 33)B1 M1 A1 [3] using the division algorithm 
12x ≡ 3 (mod 99) ⟹ 4x ≡ 1 (mod 33) ⟹ 4x = 1, 34, 67, 100, ... ⟹ x ≡ 25 (mod 33) | B1 M1 A1 | Considering equivalents mod 33; Or in the form $x = 33n + 25$ ($n \in \mathbb{Z}$); Or 25, 58, 91 (mod 99); For $4x \equiv 1$ (mod 99) B0 Then $4x = 1, 34, 67, 100, ...$ M1 (or via two steps) ⟹ $x \equiv 25$ (mod 99) or $x = 99n + 25$ ($n \in \mathbb{Z}$) A1

**or**

99 = 8×12 + 3; So -8 is a solution and hcf(99,12)=3 ⟹ x ≡ -8 (mod 33) | B1 M1 A1 [3] | using the division algorithm
1 Solve $12 x \equiv 3 ( \bmod 99 )$.

\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q1 [3]}}
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