OCR Further Additional Pure AS 2022 June — Question 4 8 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumber Theory
TypeDivisibility tests and proofs
DifficultyChallenging +1.2 This is a Further Maths number theory question requiring proof of a divisibility criterion and its application. Part (a) involves straightforward modular arithmetic manipulation (showing 10a+b ≡ 0 (mod 29) iff a+3b ≡ 0 (mod 29) using 10·3 ≡ 1 (mod 29)), which is a standard technique. Part (b) is mechanical iteration once the method is established. While this requires understanding of divisibility and modular arithmetic beyond standard A-level, it's a routine application of these techniques rather than requiring deep insight, placing it slightly above average difficulty.
Spec8.02d Division algorithm: a = bq + r uniquely8.02e Finite (modular) arithmetic: integers modulo n
4 Let \(\mathrm { N } = 10 \mathrm { a } + \mathrm { b }\) and \(\mathrm { M } = \mathrm { a } + 3 \mathrm {~b}\), where \(a\) and \(b\) are integers such that \(a \geqslant 1\) and \(0 \leqslant b \leqslant 9\).
  1. Prove that \(29 \mid N\) if and only if \(29 \mid M\).
  2. Use an iterative method based on the result of part (a) to show that 899364472 is a multiple of 29 .
Question 4:
AnswerMarks Guidance
4a 3N – M = 30a + 3b – a – 3b = 29a
Then 29N  29 M = 3N – 29a
and 29M  29 3N = M + 29a
A1
M1
A1
AnswerMarks
A13.1a
1.1
2.1
1.1
AnswerMarks
2.2aLinear combination of N and M
attempted
Multiple of 29 stated or shown
1 direction of an  proof
attempted
Shown convincingly
Including justification since
AnswerMarks
hcf(3, 29) = 1or 2N + 9M = 29(a + b)
or N + 19M = 29(a +
2b)
Alt. “or” forms above
lead to exactly similar
working; suggest the hcf
business only required
once in the 2N + 9M
case
AnswerMarks
Alternative methodM1
M1
AnswerMarks
A1First direction of  proof
Let N = 29n.
Then M = a + 3b = a + 3(N – 10a)
= 3N – 29a = 29(3n – a)
AnswerMarks Guidance
and if M = 29m then N = 10(M – 3b) + bM1 Second direction of  proof
 N = 10M – 29b = 29(10m – b)A1
[5]
AnswerMarks Guidance
b899 364 47 2 → 899 364 5
→ 899 3654 → 899 37 7
→ 899 58 → 901 9 → 92
→ 116 → 29 M1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
2.2a1st step of iterative process correct
(i.e. 10a + b replaced by a + 3b)
Method used repeatedly at least
twice correctly
Ending at 29 or at 116 = 4  29
M1
M1
A1
Question 4:
4 | a | 3N – M = 30a + 3b – a – 3b = 29a
Then 29 | N  29 | M = 3N – 29a
and 29 | M  29 | 3N = M + 29a | M1
A1
M1
A1
A1 | 3.1a
1.1
2.1
1.1
2.2a | Linear combination of N and M
attempted
Multiple of 29 stated or shown
1 direction of an  proof
attempted
Shown convincingly
Including justification since
hcf(3, 29) = 1 | or 2N + 9M = 29(a + b)
or N + 19M = 29(a +
2b)
Alt. “or” forms above
lead to exactly similar
working; suggest the hcf
business only required
once in the 2N + 9M
case
Alternative method | M1
M1
A1 | First direction of  proof
Let N = 29n.
Then M = a + 3b = a + 3(N – 10a)
= 3N – 29a = 29(3n – a)
and if M = 29m then N = 10(M – 3b) + b | M1 | Second direction of  proof
 N = 10M – 29b = 29(10m – b) | A1
[5]
b | 899 364 47 | 2 → 899 364 5 | 3
→ 899 365 | 4 → 899 37 | 7
→ 899 5 | 8 → 901 | 9 → 92 | 8
→ 11 | 6 → 29 | M1
M1
A1
[3] | 1.1
1.1
2.2a | 1st step of iterative process correct
(i.e. 10a + b replaced by a + 3b)
Method used repeatedly at least
twice correctly
Ending at 29 or at 116 = 4  29
M1
M1
A1
4 Let $\mathrm { N } = 10 \mathrm { a } + \mathrm { b }$ and $\mathrm { M } = \mathrm { a } + 3 \mathrm {~b}$, where $a$ and $b$ are integers such that $a \geqslant 1$ and $0 \leqslant b \leqslant 9$.
\begin{enumerate}[label=(\alph*)]
\item Prove that $29 \mid N$ if and only if $29 \mid M$.
\item Use an iterative method based on the result of part (a) to show that 899364472 is a multiple of 29 .
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2022 Q4 [8]}}
This paper (7 questions)
View full paper
Q1 6 Q2 6 Q3 7 Q4 8 Q5 6 Q6 14 Q7 13