OCR Further Additional Pure AS 2019 June — Question 6 8 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2019
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumber Theory
TypeLinear congruences (single)
DifficultyStandard +0.8 This is a Further Maths question on linear congruences requiring the extended Euclidean algorithm or modular inverse techniques. Part (a) is straightforward application of finding modular inverses with a prime modulus. Parts (b)(i) and (b)(ii) require recognizing that 95 ≡ -6 (mod 101) and cleverly using the result from part (a), demonstrating problem-solving beyond routine calculation. While systematic, it requires more sophistication than standard A-level content and involves multiple connected steps.
Spec8.02f Single linear congruences: solve ax = b (mod n)
6
  1. Determine all values of \(x\) for which \(16 x \equiv 5 ( \bmod 101 )\).
  2. Solve
    1. \(95 x \equiv 6 ( \bmod 101 )\),
    2. \(95 x \equiv 5 ( \bmod 101 )\).
Question 6:
AnswerMarks Guidance
6(a) DR (Working mod 101 throughout) 16x  5  106 …
 1520
Explanation that we can divide by 16 since hcf(16, 101) = 1
AnswerMarks
 x  95 (mod 101) or x = 101n + 95 or any other valid formM1
A1
E1
AnswerMarks
A11.1
2.1
2.4
AnswerMarks
2.2aAdding multiples of 101 at any stage
Finding a multiple of 16
Explained appropriately at any stage
(Use of hcf(16, 101) = 1 to justify other attributes E0)
Alternative method I Done in stages
AnswerMarks Guidance
e.g. 16x  5  106 …  8x  53M1
 8x  154  4x  77  178A1 Must be evidence of repeated divisions (correct  twice)
 2x  89  190  x  95 (mod 101) etc.A1
Explanation that we can divide by 2 since hcf(2, 101) = 1E1 Explained appropriately at any stage (once will suffice)
Alternative method II Using reciprocal/inverse
AnswerMarks
Finding 16 – 1 (mod 101) = 19M1 A1
Multiplying throughout 16x  5 (mod 101) by 19  x  95 (mod 101)M1 A1
[4]
AnswerMarks Guidance
(b)(i) DR 95x  6  –6x  6
 x  –1 (mod 101) oeM1
A13.1a
1.1
Alternative method I Using (a)
AnswerMarks Guidance
Multg. throughout by 16  5x  96 (mod 101)Or by noting that this is 5x  –5 (mod 101)
Multg. throughout by 81  405x  x  100 (mod 101)M1 A1 Complete method NB 81  5 = 405  1 (mod 101)
Alternative method II Using reciprocal/inverse
AnswerMarks Guidance
Finding 95 – 1 (mod 101) = 84 Multg. throughout by 84M1 Complete method
 x  100 (mod 101)A1 NB 84  16  81 (mod 101)
[2]
AnswerMarks Guidance
(b)(ii) Using part (a)’s answer, 95  16  5 (mod 101)
 x  16 (mod 101)M1
A1
AnswerMarks
[2]2.2a
1.1Mark may be earned by solving the linear congruence from
scratch; must be a complete method
Question 6:
6 | (a) | DR (Working mod 101 throughout) 16x  5  106 …
 1520
Explanation that we can divide by 16 since hcf(16, 101) = 1
 x  95 (mod 101) or x = 101n + 95 or any other valid form | M1
A1
E1
A1 | 1.1
2.1
2.4
2.2a | Adding multiples of 101 at any stage
Finding a multiple of 16
Explained appropriately at any stage
(Use of hcf(16, 101) = 1 to justify other attributes E0)
Alternative method I Done in stages
e.g. 16x  5  106 …  8x  53 | M1
 8x  154  4x  77  178 | A1 | Must be evidence of repeated divisions (correct  twice)
 2x  89  190  x  95 (mod 101) etc. | A1
Explanation that we can divide by 2 since hcf(2, 101) = 1 | E1 | Explained appropriately at any stage (once will suffice)
Alternative method II Using reciprocal/inverse
Finding 16 – 1 (mod 101) = 19 | M1 A1
Multiplying throughout 16x  5 (mod 101) by 19  x  95 (mod 101) | M1 A1
[4]
(b) | (i) | DR 95x  6  –6x  6
 x  –1 (mod 101) oe | M1
A1 | 3.1a
1.1
Alternative method I Using (a)
Multg. throughout by 16  5x  96 (mod 101) | Or by noting that this is 5x  –5 (mod 101)
Multg. throughout by 81  405x  x  100 (mod 101) | M1 A1 | Complete method NB 81  5 = 405  1 (mod 101)
Alternative method II Using reciprocal/inverse
Finding 95 – 1 (mod 101) = 84 Multg. throughout by 84 | M1 | Complete method
 x  100 (mod 101) | A1 | NB 84  16  81 (mod 101)
[2]
(b) | (ii) | Using part (a)’s answer, 95  16  5 (mod 101)
 x  16 (mod 101) | M1
A1
[2] | 2.2a
1.1 | Mark may be earned by solving the linear congruence from
scratch; must be a complete method
6
\begin{enumerate}[label=(\alph*)]
\item Determine all values of $x$ for which $16 x \equiv 5 ( \bmod 101 )$.
\item Solve
\begin{enumerate}[label=(\roman*)]
\item $95 x \equiv 6 ( \bmod 101 )$,
\item $95 x \equiv 5 ( \bmod 101 )$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2019 Q6 [8]}}
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