Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1997 AIME Problems/Problem 9"
m (→‎Solution 1)
Line 5: Line 5:
 
Looking at the properties of the number, it is immediately guess-able that <math>a = \phi = \frac{1+\sqrt{5}}2</math> (the [[phi|golden ratio]]) is the answer. The following is the way to derive that:
 
Looking at the properties of the number, it is immediately guess-able that <math>a = \phi = \frac{1+\sqrt{5}}2</math> (the [[phi|golden ratio]]) is the answer. The following is the way to derive that:
  
−
Since <math>\sqrt{2} < a < \sqrt{3}</math>, <math>0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1</math>. Thus <math>\langle a^2 \rangle = a^{-1}</math>, and it follows that <math>a^2 - 2 = a^{-1} \Longrightarrow a^3 - 2a - 1 = 0</math>. Noting that <math>-1</math> is a root, this factors to <math>(a+1)(a^2 - a - 1) = 0</math>, so <math>a = \frac{1 \pm \sqrt{5}}{2}</math> (we discard the negative root).
+
Since <math>\sqrt{2} < a < \sqrt{3}</math>, <math>0 < \frac{1}{\sqrt{3}} < a^{-1} < \frac{1}{\sqrt{2}} < 1</math>. Thus <math>\langle a^2 \rangle = a^{-1}</math>, and it follows that <math>a^2 - 2 = a^{-1} \Longrightarrow a^3 - 2a - 1 = 0</math>. Noting that <math>-1</math> is a root, this factors to <math>(a+1)(a^2 - a - 1) = 0</math>, so <math>a = \frac{1 + \sqrt{5}}{2}</math> (we discard the negative root).
  
 
Our answer is <math>(a^2)^{6}-144a^{-1} = \left(\frac{3+\sqrt{5}}2\right)^6 - 144\left(\frac{2}{1 + \sqrt{5}}\right)</math>. [[Complex conjugate]]s reduce the second term to <math>-72(\sqrt{5}-1)</math>. The first term we can expand by the [[binomial theorem]] to get <math>\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)</math> <math>= \frac{1}{64}\left(10304 + 4608\sqrt{5}\right) = 161 + 72\sqrt{5}</math>. The answer is <math>161 + 72\sqrt{5} - 72\sqrt{5} + 72 = \boxed{233}</math>.
 
Our answer is <math>(a^2)^{6}-144a^{-1} = \left(\frac{3+\sqrt{5}}2\right)^6 - 144\left(\frac{2}{1 + \sqrt{5}}\right)</math>. [[Complex conjugate]]s reduce the second term to <math>-72(\sqrt{5}-1)</math>. The first term we can expand by the [[binomial theorem]] to get <math>\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)</math> <math>= \frac{1}{64}\left(10304 + 4608\sqrt{5}\right) = 161 + 72\sqrt{5}</math>. The answer is <math>161 + 72\sqrt{5} - 72\sqrt{5} + 72 = \boxed{233}</math>.

Revision as of 18:16, 27 July 2020

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 1

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 $a = \frac{1 + \sqrt{5}}{2}$ (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 $\phi^3 = 2\phi + 1$.

Solution 2

Find $a$ as shown above. Note that, since $a$ is a root of the equation $a^3 - 2a - 1 = 0$, $a^3 = 2a + 1$, and $a^{12} = (2a + 1)^4$. Also note that, since $a$ is a root of $a^2 - a - 1 = 0$, $\frac{1}{a} = a - 1$. The expression we wish to calculate then becomes $(2a + 1)^4 - 144(a - 1)$. Plugging in $a = \frac{1 + \sqrt{5}}{2}$, we plug in to get an answer of $(161 + 72\sqrt{5}) + 72 - 72\sqrt{5} = 161 + 72 = \boxed{233}$.

Solution 3

Find $a$ as shown above. Note that $a$ satisfies the equation $a^2 = a+1$ (this is the equation we solved to get it). Then, we can simplify $a^{12}$ as follows using the fibonacci numbers:

$a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^8+ 2a^9 = ... = 144a^1+89a^0 = 144a+89$

So we want $144(a-\frac1a)+89 = 144(1)+89 = \boxed{233}$ since $a-\frac1a = 1$ is equivalent to $a^2 = a+1$.

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1997_AIME_Problems/Problem_9&oldid=129557"