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Old 28 Jan 2008, 10:55 AM   #1 (permalink)
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Maths Problems for a thought

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If you've entered this section, you must love maths. Why not share the love for maths?


Post interesting maths problem solving questions. =D
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Old 29 Jan 2008, 11:33 AM   #2 (permalink)
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Re: Maths Problems for a thought

let a = b.
multiply both sides by a.
a^2 = ab
add a^2 to both sides.
2a^2 = a^2 + ab
subtract 2ab from both sides.
2a^2 - 2ab = a^2 - ab
take out a^2 - ab as a common factor.
2(a^2 - ab) = 1(a^2 - ab)
divide through by a^2 - ab.
2 = 1
subtract 1 from both sides.
1 = 0

QED.
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Old 29 Jan 2008, 11:38 AM   #3 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by humphdogg
let a = b.
multiply both sides by a.
a^2 = ab
add a^2 to both sides.
2a^2 = a^2 + ab
subtract 2ab from both sides.
2a^2 - 2ab = a^2 - ab
take out a^2 - ab as a common factor.
2(a^2 - ab) = 1(a^2 - ab)
divide through by a^2 - ab.
2 = 1
subtract 1 from both sides.
1 = 0

QED.
but you've said a^2 = ab ---> a^2 - ab = 0 hence you cant do that step.
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Old 29 Jan 2008, 2:04 PM   #4 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by watatank
but you've said a^2 = ab ---> a^2 - ab = 0 hence you cant do that step.
He knows that :P
You can't prove 0=1
Humphdogg was just mucking around.
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Old 20 Mar 2008, 8:45 PM   #5 (permalink)
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Re: Maths Problems for a thought

0.9999999999999 = 1
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Old 22 Mar 2008, 4:17 PM   #6 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by Roguedeth
0.9999999999999 = 1

¬¬

0.999999999 ~ 1
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Old 22 Mar 2008, 5:49 PM   #7 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by watatank
but you've said a^2 = ab ---> a^2 - ab = 0 hence you cant do that step.
well actually since he is using two pronumerals, he can do that step...i know, i know, pointless statement for a pointless arugment.
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Old 29 May 2008, 8:19 AM   #8 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by 3unitz
i like this integral
dx / (x^3 + 2x + 3)

(x^4 / 4) + [(2x^2)/2] + 3x + C

= ( 45 surd 5) /8 + (111/ 8) + C

Is that correct?
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Old 29 May 2008, 8:42 AM   #9 (permalink)
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Re: Maths Problems for a thought

No, try partial fractions
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Old 13 Jul 2008, 1:44 PM   #10 (permalink)
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Post Re: Maths Problems for a thought

what is the last digit of this number?

333^444

(a) 1
(b) 3
(c) 7
(d) 9
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Old 13 Jul 2008, 3:06 PM   #11 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by denominator
what is the last digit of this number?

333^444

(a) 1
(b) 3
(c) 7
(d) 9
a) 1.

it repeats every 4 powers with 1 being the last digit of 333^0
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Old 13 Jul 2008, 3:15 PM   #12 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by aMUSEd1977
Prove 0 = 1
I can't.

Quote:
Originally Posted by denominator
what is the last digit of this number?

333^444

(a) 1
(b) 3
(c) 7
(d) 9
None. It's a maths error.

I am joking. It's 1 as lolokay already said.
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Old 13 Jul 2008, 3:20 PM   #13 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by lyounamu
None. It's a maths error.
lol
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Old 1 Nov 2008, 3:37 PM   #14 (permalink)
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Re: Maths Problems for a thought

Actually 0.9' does = 1

Let x = 0.9'
Therefor 10x = 9.9'
Therefor 9x= 9
Therefore x = 1
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Old 2 Nov 2008, 10:26 AM   #15 (permalink)
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Re: Maths Problems for a thought

Quote:
Originally Posted by field5
Actually 0.9' does = 1

Let x = 0.9'
Therefor 10x = 9.9'
Therefor 9x= 9
Therefore x = 1
How can you subtract 'x' from either side if 'x' is only on one side?
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