| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Topic | Proof by induction |
| Type | Prove divisibility |
| Difficulty | Challenging +1.8 This is a non-standard induction proof requiring students to discover the pattern in products like 13579×11111, establish a recurrence relation between consecutive cases, and carefully track how the string of 7's grows. The novelty of repunit numbers and the need to manipulate decimal representations makes this significantly harder than routine induction proofs on summation formulas, though the inductive step follows standard structure once the pattern is identified. |
| Spec | 4.01a Mathematical induction: construct proofs |
Base-line case: for $n = 5$, $13579\,R_5 = 15087\,6269$ contains a string of $(5-4=1)$ 7's **B1**
$13579\,R_6 = 1508776269$, $13579\,R_7 = 150877766269$, etc. or form of 1st & last 4 digits **B1**
Assume that, for some $k \geq 5$, $13579\,R_k = 1508\underbrace{77\ldots7}_{(k-4)\,7\text{'s}}6269$. 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.
$= 1508\underbrace{77\ldots7}_{(k-4)\,7\text{'s}}62690$
$\quad + 13579$
$= 1508\underbrace{77\ldots7}_{(k-4+1)\,7\text{'s}}76269$ **A1**
which contains a string of $(k - 4 + 1)$ 7's, as required. Proof follows by induction (usual round-up). **E1**
**[6]**13 Define the repunit number, $R _ { n }$, to be the positive integer which consists of a string of $n 1$ '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 7's.
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2012 Q13 [6]}}