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 Post subject: Equation
PostPosted: Sun, 15 Jan 2012 15:13:22 UTC 
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Is given

$m=\frac{2^a+6^a+8^a}{1+3^a+4^a}\;\;\;,\;\;a\in N\;\;\text{which number below can be m?}


$A)\;18\;\;\;B)\;72\;\;\;C)\;100\;\;D)\;256


$m=\frac{2^a+6^a+8^a}{1+3^a+4^a}=\frac{2^a(1+3^a+4^a)}{1+3^a+4^a}=2^a\;\;=>\;m=2^a

\text{If:}

$a=1\;\;=>\;m=2

$a=2\;\;=>\;m=4

$a=3\;\;=>\;m=8

$a=4\;\;=>\;m=16

$a=5\;\;=>\;m=32

$a=6\;\;=>\;m=64

$a=7\;\;=>\;m=128

$a=8\;\;=>\;m=256

$a=9\;\;=>\;m=512

...if I'm right, D should be the correct answer. But I want to know if there's any other way of doing this or not. :(
Thank you.


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 Post subject: Re: Equation
PostPosted: Sun, 15 Jan 2012 15:30:46 UTC 
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log(18) / log(2) = ~4.17
log(72) / log(2) = ~6.17
log(100)/ log(2) = ~6.64
log(256)/ log(2) = 8.000000000000000000000000000000000000000000.......

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 Post subject: Re: Equation
PostPosted: Sun, 15 Jan 2012 16:53:12 UTC 
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Denis wrote:
log(18) / log(2) = ~4.17
log(72) / log(2) = ~6.17
log(100)/ log(2) = ~6.64
log(256)/ log(2) = 8.000000000000000000000000000000000000000000.......


:mrgreen: I guess you're angry with those zeros about something. :!:

You know what, I was thinking about your advice, wristwatch should be perfect, but only a temporary one, with ball pen :lol:
Thanks.


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 Post subject: Re: Equation
PostPosted: Sun, 15 Jan 2012 23:51:55 UTC 
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kreshnik wrote:
Denis wrote:
log(18) / log(2) = ~4.17
log(72) / log(2) = ~6.17
log(100)/ log(2) = ~6.64
log(256)/ log(2) = 8.000000000000000000000000000000000000000000.......


:mrgreen: I guess you're angry with those zeros about something. :!:

You know what, I was thinking about your advice, wristwatch should be perfect, but only a temporary one, with ball pen :lol:
Thanks.


Really, small perfect powers (e.g. those below 1000), or failing that, small perfect prime powers, should be memorised.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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