Difference between revisions of "2005 AIME II Problems/Problem 7"

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== Problem ==
 
== Problem ==
Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math> (x+1)^{48}. </math>
 
  
== Solution ==
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Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>(x+1)^{48}</math>.
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== Solution 1==
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We note that in general,
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<center>
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<math>{} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>.
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</center>
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It now becomes apparent that if we multiply the [[numerator]] and [[denominator]] of <math>\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) } </math> by <math>(\sqrt[16]{5} - 1) </math>, the denominator will [[telescope]] to <math>\sqrt[1]{5} - 1 = 4 </math>, so
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<center>
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<math>x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1</math>.
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</center>
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It follows that <math>(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}</math>.
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== Solution 2 (Bashing) ==
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Let <math>y=\sqrt[16]{5}</math>, then expanding the denominator results in:
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<cmath>(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)</cmath>
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<cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath>
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<cmath>(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>
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Therefore:
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<cmath>\frac{4}{(y^{16}-1)/(y-1)} = \frac{4(y-1)}{y^{16}-1}=\frac{4(y-1)}{4} = y-1 </cmath>
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It is evident that <math> x+1 = (y-1)+1 = \sqrt[16]5</math>  as Solution 1 states.
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== Solution 3 ==
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Like Solution <math>2</math>, let <math>z=\sqrt[16]{5}</math>
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Then, the expression becomes
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<math>x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}</math>
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Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
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<math>x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}</math>
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Plugging back in,
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<math>x=\frac{4({\sqrt[16]{5}-1)}}{5-1}\implies x=\sqrt[16]{5}-1</math>
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Hence, after some basic exponent rules, we find the answer is <math>\boxed{125}</math>
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== See Also ==
 
== See Also ==
*[[2005 AIME II Problems]]
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{{AIME box|year=2005|n=II|num-b=6|num-a=8}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:33, 27 October 2019

Problem

Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$.

Solution 1

We note that in general,

${} (\sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1$.

It now becomes apparent that if we multiply the numerator and denominator of $\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $(\sqrt[16]{5} - 1)$, the denominator will telescope to $\sqrt[1]{5} - 1 = 4$, so

$x = \frac{4(\sqrt[16]{5} - 1)}{4} = \sqrt[16]{5} - 1$.

It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}$.

Solution 2 (Bashing)

Let $y=\sqrt[16]{5}$, then expanding the denominator results in: \[(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)\] \[(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)\] \[(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}\]

Therefore: \[\frac{4}{(y^{16}-1)/(y-1)} = \frac{4(y-1)}{y^{16}-1}=\frac{4(y-1)}{4} = y-1\]

It is evident that $x+1 = (y-1)+1 = \sqrt[16]5$ as Solution 1 states.

Solution 3

Like Solution $2$, let $z=\sqrt[16]{5}$ Then, the expression becomes

$x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}$ Now, multiplying by the conjugate of each binomial in the denominator, we obtain...

$x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}$ Plugging back in,

$x=\frac{4({\sqrt[16]{5}-1)}}{5-1}\implies x=\sqrt[16]{5}-1$

Hence, after some basic exponent rules, we find the answer is $\boxed{125}$

See Also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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