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Difference between revisions of "1997 AIME Problems/Problem 9"

 
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== Problem ==
 
== Problem ==
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Given a [[nonnegative]] real number <math>x</math>, let <math>\langle x\rangle</math> denote the fractional part of <math>x</math>; that is, <math>\langle x\rangle=x-\lfloor x\rfloor</math>, where <math>\lfloor x\rfloor</math> denotes the [[greatest integer]] less than or equal to <math>x</math>. Suppose that <math>a</math> is positive, <math>\langle a^{-1}\rangle=\langle a^2\rangle</math>, and <math>2<a^2<3</math>. Find the value of <math>a^{12}-144a^{-1}</math>.
  
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== Solution ==
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== Solution 1==
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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:
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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).
 +
 
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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>.
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Note that to determine our answer, we could have also used other properties of <math>\phi</math> like <math>\phi^3 = 2\phi + 1</math>.
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== Solution 2 ==
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Find <math>a</math> as shown above. Note that, since <math>a</math> is a root of the equation <math>a^3 - 2a - 1 = 0</math>, <math>a^3 = 2a + 1</math>, and <math>a^{12} = (2a + 1)^4</math>. Also note that, since <math>a</math> is a root of <math>a^2 - a - 1 = 0</math>, <math>\frac{1}{a} = a - 1</math>. The expression we wish to calculate then becomes <math>(2a + 1)^4 - 144(a - 1)</math>. Plugging in <math>a = \frac{1 + \sqrt{5}}{2}</math>, we plug in to get an answer of <math>(161 + 72\sqrt{5}) + 72 - 72\sqrt{5} = 161 + 72 = \boxed{233}</math>.
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== Solution 3 ==
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Find <math>a</math> as shown above. Note that <math>a</math> satisfies the equation <math>a^2 = a+1</math> (this is the equation we solved to get it). Then, we can simplify <math>a^{12}</math> as follows using the fibonacci numbers:
 +
 
 +
<math>a^{12} = a^{11}+a^{10}= 2a^{10} + a^{9} = 3a^9+ 2a^8 = ... = 144a^1+89a^0 = 144a+89</math>
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So we want <math>144(a-\frac1a)+89 = 144(1)+89 = \boxed{233}</math> since <math>a-\frac1a = 1</math> is equivalent to <math>a^2 = a+1</math>.
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==Solution 4==
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As Solution 1 stated, <math>a^3 - 2a - 1 = 0</math>. <math>a^3 - 2a - 1 = a^3 - a^2 -a + a^2 -a -1 = (a+1)(a^2 - a - 1)</math>. So, <math>a^2 - a - 1 = 0</math>, <math>1 = a^2 - a</math>, <math>\frac1a = a-1</math>, <math>a^3 = 2a+1</math>, <math>a^2 = a+1</math>.
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<math>a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5</math>
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<math>a^{12} = (a^6)^2 = (8a+5)^2 = 64a^2 + 80a + 25 = 64(a+1) + 80a + 25 = 144a + 89</math>
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Therefore, <math>a^{12} - 144 a^{-1} = 144a + 89 - 144(a-1) = 89 + 144 = \boxed{233}</math>
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Another way to factor <math>a^3 - 2a - 1</math>:
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<math>a^3 - 2a - 1 = a^3 + 1 -2a -2 = (a+1)(a^2 - a + 1) - 2(a+1) = (a + 1)(a^2 - a - 1)</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== See also ==
 
== See also ==
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* [[1997 AIME Problems]]
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{{AIME box|year=1997|num-b=8|num-a=10}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:53, 30 December 2023

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:

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

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

Solution 4

As Solution 1 stated, $a^3 - 2a - 1 = 0$. $a^3 - 2a - 1 = a^3 - a^2 -a + a^2 -a -1 = (a+1)(a^2 - a - 1)$. So, $a^2 - a - 1 = 0$, $1 = a^2 - a$, $\frac1a = a-1$, $a^3 = 2a+1$, $a^2 = a+1$.

$a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5$

$a^{12} = (a^6)^2 = (8a+5)^2 = 64a^2 + 80a + 25 = 64(a+1) + 80a + 25 = 144a + 89$

Therefore, $a^{12} - 144 a^{-1} = 144a + 89 - 144(a-1) = 89 + 144 = \boxed{233}$


Another way to factor $a^3 - 2a - 1$:

$a^3 - 2a - 1 = a^3 + 1 -2a -2 = (a+1)(a^2 - a + 1) - 2(a+1) = (a + 1)(a^2 - a - 1)$

~isabelchen

See also

1997 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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