1997 AIME Problems/Problem 9

From AoPSWiki

Problem

Given a nonnegative real number $x$ , let $\langle x\rangle$ denote the fractional part of $x$ ; that is, $\langle x\rangle=x-\lfloor x\rfloor$ , where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ . Suppose that $a$ is positive, $\langle a^{-1}\rangle=\langle a^2\rangle$ , and $2<a^2<3$ . Find the value of $a^{12}-144a^{-1}$ .

Solution

Looking at the properties of the number, it is immediately guess-able that $a = \phi = \frac{1+\sqrt{5}}2$ (the golden ratio) is the answer. The following is the way to derive that:

Since $\sqrt{2} < a < \sqrt{3}$ , $0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1$ . Thus $\langle a^2 \rangle = a^{-1}$ , and it follows that $a^2 - 2 = a^{-1} \Longrightarrow a^3 - 2a - 1 = 0$ . Noting that $-1$ is a root, this factors to $(a+1)(a^2 - a - 1) = 0$ , so $a = \frac{1 \pm \sqrt{5}}{2}$ (we discard the negative root).

Our answer is $(a^2)^{6}-144a^{-1} = \left(\frac{3+\sqrt{5}}2\right)^6 - 144\left(\frac{2}{1 + \sqrt{5}}\right)$ . Complex conjugates reduce the second term to $-72(\sqrt{5}-1)$ . The first term we can expand by the binomial theorem to get $\frac 1{2^6}\left(3^6 + 6\cdot 3^5\sqrt{5} + 15\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)$ $= \frac{1}{64}\left(10304 + 4608\sqrt{5}\right) = 161 + 72\sqrt{5}$ . The answer is $161 + 72\sqrt{5} - 72\sqrt{5} + 72 = \boxed{233}$ .

Note that to determine our answer, we could have also used other properties of $\phi$ like $\phi^3 = 2\phi + 1$ .

1997 AIME (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

1997 AIME Problems/Problem 9

From AoPSWiki

Problem

Solution

See also