 October 05, 2023, 01:05:59 am

### AuthorTopic: Mathematics Extension 1 Challenge Marathon  (Read 25276 times) Tweet Share

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#### Opengangs

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« Reply #45 on: December 29, 2018, 02:14:35 am »
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$a) \text{ Show that } \\ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$
$\text{Let } p(x) = x^3 + 12x^2 - 2x + 1 \text{ have roots } \alpha, \beta, \gamma. \\ b) \text{ Explain why} \\ \alpha^3 = 2\alpha - 12\alpha^2 - 1.$$c) \text{ Hence, or otherwise, find } \alpha^3 + \beta^3 + \gamma^3.$

#### david.wang28

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« Reply #46 on: December 30, 2018, 11:41:07 pm »
+1
$a) \text{ Show that } \\ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)$
$\text{Let } p(x) = x^3 + 12x^2 - 2x + 1 \text{ have roots } \alpha, \beta, \gamma. \\ b) \text{ Explain why} \\ \alpha^3 = 2\alpha - 12\alpha^2 - 1.$$c) \text{ Hence, or otherwise, find } \alpha^3 + \beta^3 + \gamma^3.$
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#### Opengangs

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« Reply #47 on: February 05, 2019, 02:45:56 pm »
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Not so much of a "challenge" but an interesting proof to a property you guys should be familiar with.
« Last Edit: February 05, 2019, 02:50:00 pm by Opengangs »

#### Opengangs

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« Reply #48 on: May 14, 2020, 10:55:24 pm »
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A fun (and not too hard) problem that ties into number theory a little :-))

Our goal here is to investigate $n^5 - n$ and show that, indeed it is divisible by 5. We will verify this is true using induction (that's for you to prove) at the end. But for now, let's explore this using some number theory.

Part 1: Show that $n^5 - n$ can be written as $n(n - 1)(n + 1)(n^2 + 1)$.
Part 2: If we choose $n = 5k$, $n = 5k + 1$ or $n = 5k + 4$, briefly explain why $n^5 - n$ is divisible by 5 for any integer $k$.
Part 3: Show that, if we choose $n = 5k + 2$ or $n = 5k + 3$, then $n^5 - n$ is also divisible by 5.
Part 4: Conclude that, regardless of our choice of $n$, $n^5 - n$ is always divisible by 5.
Part 5: Verify this statement using induction. That is, prove that $n^5 - n$ is divisible by 5 using mathematical induction.