Pre-U Pre-U 9795/1 2016 Specimen — Question 13 6 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2016
SessionSpecimen
Marks6
TopicNumber Theory
TypeModular arithmetic properties
DifficultyChallenging +1.8 This is a challenging Further Maths induction proof requiring students to discover the pattern through multiplication, establish a non-trivial base case (n=5), and execute a sophisticated inductive step involving algebraic manipulation of repunits. The conceptual leap of recognizing how the string of 7s propagates is substantial, placing this well above standard induction proofs but not at the extreme difficulty of olympiad-style number theory.
Spec4.01a Mathematical induction: construct proofs8.02a Number bases: conversion and arithmetic in base n

13 Define the repunit number, \(R _ { n }\), to be the positive integer which consists of a string of \(n 1 \mathrm {~s}\). Thus, $$R _ { 1 } = 1 , \quad R _ { 2 } = 11 , \quad R _ { 3 } = 111 , \quad \ldots , \quad R _ { 7 } = 1111111 , \quad \ldots , \text { etc. }$$ Use induction to prove that, for all integers \(n \geqslant 5\), the number $$13579 \times R _ { n }$$ contains a string of ( \(n - 4\) ) consecutive 7s.

Base-line case: for \(n = 5\), \(13579\,R_5 = 1508\,7\,6269\) contains a string of \((5-4=1)\) 7s B1
\(13579\,R_6 = 1508776269\), \(13579\,R_7 = 15087776269\), etc. or form of 1st & last 4 digits B1
Assume that, for some \(k \geq 5\), \(13579\,R_k = 1508\underbrace{77\ldots7}_{(k-4)\text{ 7's}}6269\) — Induction hypothesis M1
Then, for \(n = k+1\),
\(13579\,R_{k+1} = 13579(10R_k + 1)\) M1
Give the M mark for the key observation that \(R_{k+1} = 10R_k + 1\) or \(10^k + R_k\), even if not subsequently used.
\(= 1508\underbrace{77\ldots7}_{(k-4)\text{ 7's}}62690\)
\(\quad + 13579\)
\(= 1508\underbrace{77\ldots7}_{(k-4+1)\text{ 7's}}76269\) A1
which contains a string of \((k-4+1)\) 7s, as required. Proof follows by induction (usual round-up). A1
Total: 6 marks
Base-line case: for $n = 5$, $13579\,R_5 = 1508\,7\,6269$ contains a string of $(5-4=1)$ 7s **B1**

$13579\,R_6 = 1508776269$, $13579\,R_7 = 15087776269$, etc. or form of 1st & last 4 digits **B1**

Assume that, for some $k \geq 5$, $13579\,R_k = 1508\underbrace{77\ldots7}_{(k-4)\text{ 7's}}6269$ — Induction hypothesis **M1**

Then, for $n = k+1$,
$13579\,R_{k+1} = 13579(10R_k + 1)$ **M1**

Give the **M** mark for the key observation that $R_{k+1} = 10R_k + 1$ **or** $10^k + R_k$, even if not subsequently used.

$= 1508\underbrace{77\ldots7}_{(k-4)\text{ 7's}}62690$

$\quad + 13579$

$= 1508\underbrace{77\ldots7}_{(k-4+1)\text{ 7's}}76269$ **A1**

which contains a string of $(k-4+1)$ 7s, as required. Proof follows by induction (usual round-up). **A1**

**Total: 6 marks**
13 Define the repunit number, $R _ { n }$, to be the positive integer which consists of a string of $n 1 \mathrm {~s}$. Thus,

$$R _ { 1 } = 1 , \quad R _ { 2 } = 11 , \quad R _ { 3 } = 111 , \quad \ldots , \quad R _ { 7 } = 1111111 , \quad \ldots , \text { etc. }$$

Use induction to prove that, for all integers $n \geqslant 5$, the number

$$13579 \times R _ { n }$$

contains a string of ( $n - 4$ ) consecutive 7s.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2016 Q13 [6]}}