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January 29, 2020, 03:42:41 am

Author Topic: Recreational Problems (MM level)  (Read 12997 times)  Share 

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Flaming_Arrow

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Recreational Problems (MM level)
« on: June 12, 2008, 09:33:08 pm »
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btw im in yr 11 so this is question from unit 1 and 2 book

ill start off




« Last Edit: June 12, 2008, 10:55:49 pm by chathuranj »
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Ahmad

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« Reply #1 on: June 12, 2008, 09:46:08 pm »
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Stickied!
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AppleXY

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« Reply #2 on: June 12, 2008, 09:59:15 pm »
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uhm? Ok.... it's not hard lol.

solve for x and sub in.









sub in to



etc etc,

x = 79 for +ve and -ve :)


MOD EDIT: prav you need to use "\frac" and "\sqrt", and +- is "\pm" :)
« Last Edit: June 12, 2008, 10:10:52 pm by AppleXY »

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[quote="Benjamin F

unknown id

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« Reply #3 on: June 12, 2008, 10:10:48 pm »
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VCE Outline:
2007:   Accounting [48]

2008:   English [44], Maths Methods [50], Specialist Maths [41], Chemistry [50], Physics [44]

ENTER: 99.70





Flaming_Arrow

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« Reply #4 on: June 12, 2008, 10:56:19 pm »
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yep solved by applexy and unknown id
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Flaming_Arrow

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« Reply #5 on: June 14, 2008, 11:36:34 am »
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come on guys post some questions  :-[
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Ahmad

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« Reply #6 on: June 14, 2008, 12:32:52 pm »
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Show that for is never the square of an integer.
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Mao

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Recreational Problems (MM level)
« Reply #7 on: June 14, 2008, 12:48:54 pm »
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Show that for is never the square of an integer.

[IGNORE ALL THIS, COMPLETELY WRONG :P ]

in order for this to be a square, this function need to be able to be represented as for a particular value of n

initial factorising gives



cannot be further factorised [without going into imaginary, hence for that expression to represent a square of an integer,

, NOT a positive integer.

hence, for , is never the square of an integer
« Last Edit: June 14, 2008, 07:00:11 pm by Mao »
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Ahmad

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« Reply #8 on: June 14, 2008, 12:56:13 pm »
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Hmm, Mao, cannot be expressed as (without leaving the integers) yet it is a square when .

If you prove that an expression cannot be expressed as a square, it means it can not be a square for every x. But it's still possible that it's a square for a particular value of x.



I also thought I'd add 1 more problem,

Let be the familiar fibonacci numbers (with ).

Find

Back to chemistry study for me :P
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Neobeo

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« Reply #9 on: June 14, 2008, 01:15:14 pm »
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Let be the familiar fibonacci numbers (with ).

Find

Let

Then


Giving us


Also




I should come up with a new problem shortly. But study comes first!
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bigtick

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« Reply #10 on: June 14, 2008, 05:39:36 pm »
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=(n^2+an+1)^2, no rational a will satisfy
« Last Edit: June 14, 2008, 05:42:48 pm by bigtick »

Mao

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« Reply #11 on: June 14, 2008, 06:20:15 pm »
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Hmm, Mao, cannot be expressed as (without leaving the integers) yet it is a square when .

If you prove that an expression cannot be expressed as a square, it means it can not be a square for every x. But it's still possible that it's a square for a particular value of x.


but i did try to prove for a particular value of n! :P

[my basic principle is, if it is a perfect square for a particular value of n, then at some point it must be able to be expressed as a perfect square]

that is, at some point, , which is false for integer values of n.
« Last Edit: June 14, 2008, 06:23:48 pm by Mao »
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Neobeo

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« Reply #12 on: June 14, 2008, 06:24:37 pm »
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, NOT a positive integer.

You don't necessarily have to multiply two equal numbers to get a square.
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Mao

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« Reply #13 on: June 14, 2008, 07:07:34 pm »
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IN HINDSIGHT (thanks to dcc):



let this be a perfect square:



where k is an integer

hence, we must require to be rational.

but if we look at , , with the difference between the terms increasing. hence will not be a perfect square for the designated set of n, and will be irrational for

hence, it cannot be a perfect square


[i think that's what dcc meant :P ]
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« Reply #14 on: June 14, 2008, 07:19:42 pm »
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find such that the above system has non-zero solutions for
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