Difference between revisions of "1997 AIME Problems/Problem 9"
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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>. | 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>. | ||
− | == Solution == | + | == Solution 1== |
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: | ||
Revision as of 23:53, 15 March 2011
Contents
[hide]Problem
Given a nonnegative real number , let
denote the fractional part of
; that is,
, where
denotes the greatest integer less than or equal to
. Suppose that
is positive,
, and
. Find the value of
.
Solution 1
Looking at the properties of the number, it is immediately guess-able that (the golden ratio) is the answer. The following is the way to derive that:
Since ,
. Thus
, and it follows that
. Noting that
is a root, this factors to
, so
(we discard the negative root).
Our answer is . Complex conjugates reduce the second term to
. The first term we can expand by the binomial theorem to get
. The answer is
.
Note that to determine our answer, we could have also used other properties of like
.
Solution 2
Find as shown above. Note that
satisfies the equation
(this is the equation we solved to get it). Then, we can simplify
as follows using the fibonacci numbers:
So we want since
is equivalent to
.
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 |