Difference between revisions of "1997 AIME Problems/Problem 9"
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==Solution 4== | ==Solution 4== | ||
− | 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 | + | 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>. |
<math>a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5</math> | <math>a^6 = (a^3)^2 = (2a+1)^2= 4a^2 + 4a +1= 4(a+1) + 4a + 1= 8a+5</math> |
Revision as of 08:48, 29 March 2022
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, since
is a root of the equation
,
, and
. Also note that, since
is a root of
,
. The expression we wish to calculate then becomes
. Plugging in
, we plug in to get an answer of
.
Solution 3
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
.
Solution 4
As Solution 1 stated, .
. So,
,
,
,
,
.
Therefore,
Another way to factor :
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.