Difference between revisions of "1988 AIME Problems/Problem 13"

Revision as of 17:55, 28 September 2007

Problem

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ .

Solution

Solution 1

Let's work backwards! Let $F(x) = ax^{17} + bx^{16} + 1$ and let $P(x)$ be the polynomial such that $P(x)(x^2 - x - 1) = F(x)$ .

First, it's kinda obvious that the constant term of $P(x)$ must be $- 1$ . Now, we have $(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1)$ , where $c_{15}$ is some random coefficient. However, since $F(x)$ has no $x$ term, it must be true that $c_{15} = - 1$ .

Let's find $c_{14}$ now. Notice that all we care about in finding $c_{14}$ is that $(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}$ . Therefore, $c_{14} = - 2$ . Undergoing a similar process, $c_{13} = 3$ , $c_{12} = - 5$ , $c_{11} = 8$ , and we see a nice pattern. The coefficients of $P(x)$ are just the Fibonacci sequence with alternating signs! Therefore, $a = c_1 = F_{16}$ , where $F_{16}$ denotes the 16th Fibonnaci number and $a = 987$ .

Solution 2

Let $F_n$ represent the $n$ th number in the Fibonacci sequence. Therefore,

$x^2 - x - 1 = 0\Longrightarrow x^n = F_n(x),\ n\in N\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N\ .$

The above uses the similarity between the Fibonacci recursive definition, $F_{n+2} - F_{n+1} - F_n = 0$ , and the polynomial $x^2 - x - 1 = 0$ .

$0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow$

$(aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q\Longrightarrow$

$aF_{17} + bF_{16} = 0$ and $aF_{16} + bF_{15} + 1 = 0\Longrightarrow$

$a = F_{16},\ b = - F_{17}\Longrightarrow \boxed {a = 987,\ b = - 1597}\ .$

Solution 3

We can long divide and search for a pattern; then the remainder would be set to zero to solve for $a$ . Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is $(F_{16} + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0$ . Since the coefficient of $x$ must be zero, this gives us two equations, $F_{16}b + F_{17}a = 0$ and $F_{15}b + F_{16}a + 1 = 0$ . Solving these two as above, we get that $a = 987$ .

There are various similar solutions which yield the same pattern, such as repeated substitution of $x^2 = x + 1$ into the larger polynomial.

@@ Line 1: / Line 1: @@
 == Problem ==
+Find <math>a</math> if <math>a</math> and <math>b</math> are [[integer]]s such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>.
+__TOC__
 == Solution ==
+=== Solution 1 ===
+Let's work backwards! Let <math>F(x) = ax^{17} + bx^{16} + 1</math> and let <math>P(x)</math> be the [[polynomial]] such that <math>P(x)(x^2 - x - 1) = F(x)</math>.
+First, it's kinda obvious that the [[constant]] term of <math>P(x)</math> must be <math>- 1</math>. Now, we have <math>(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1)</math>, where <math>c_{15}</math> is some random [[coefficient]]. However, since <math>F(x)</math> has no <math>x</math> term, it must be true that <math>c_{15} = - 1</math>.
+Let's find <math>c_{14}</math> now. Notice that all we care about in finding <math>c_{14}</math> is that <math>(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}</math>. Therefore, <math>c_{14} = - 2</math>. Undergoing a similar process, <math>c_{13} = 3</math>, <math>c_{12} = - 5</math>, <math>c_{11} = 8</math>, and we see a nice pattern. The coefficients of <math>P(x)</math> are just the [[Fibonacci sequence]] with alternating signs! Therefore, <math>a = c_1 = F_{16}</math>, where <math>F_{16}</math> denotes the 16th Fibonnaci number and <math>a = 987</math>.
+=== Solution 2 ===
+Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore,
+<math>x^2 - x - 1 = 0\Longrightarrow x^n = F_n(x),\ n\in N\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N\ .</math>
+The above uses the similarity between the Fibonacci [[recursion|recursive]] definition, <math>F_{n+2} - F_{n+1} - F_n = 0</math>, and the polynomial <math>x^2 - x - 1 = 0</math>.
+<math>0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow</math>
+<math>(aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q\Longrightarrow</math>
+<math>aF_{17} + bF_{16} = 0</math> and <math>aF_{16} + bF_{15} + 1 = 0\Longrightarrow</math>
+<math>a = F_{16},\ b = - F_{17}\Longrightarrow \boxed {a = 987,\ b = - 1597}\ .</math>
+=== Solution 3 ===
+We can long divide and search for a pattern; then the remainder would be set to zero to solve for <math>a</math>. Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is <math>(F_{16} + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0</math>. Since the coefficient of <math>x</math> must be zero, this gives us two equations, <math>F_{16}b + F_{17}a = 0</math> and <math>F_{15}b + F_{16}a + 1 = 0</math>. Solving these two as above, we get that <math>a = 987</math>.
+There are various similar solutions which yield the same pattern, such as repeated substitution of <math>x^2 = x + 1</math> into the larger polynomial.
 == See also ==
-* [[1988 AIME Problems]]
+{{AIME box|year=1988|num-b=12|num-a=14}}
-{{AIME box|year=1988|num-b=12|num-a=14}}
+[[Category:Intermediate Algebra Problems]]

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