2005 AIME II Problems/Problem 7

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 $\boxed{125}$

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2005 AIME II Problems/Problem 7

Contents

Problem

Solution 1

Solution 2 (Bashing)

Solution 3

See Also