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

(Another more easier solution for beginners to AIME problems, utilizing the fibonacci pattern.)
(I fixed some of the LaTeX errors in this page. I have not intended to misrepresent the wording of the author -- please let me know if I did so. Thank you!)
Line 34: Line 34:
  
 
=== Solution 5: For Beginners (less technical) ===
 
=== Solution 5: For Beginners (less technical) ===
−
Trying to divide <math>ax^17 + bx^16 + 1</math> by <math>x^2-x-1</math> would be very tough, so let's try to divide using smaller degrees of x. Doing <math>\frac{ax^3+bx^2+1}{x^2-x-1}</math>, we get the following systems of equations:
+
Trying to divide <math>ax^{17} + bx^{16} + 1</math> by <math>x^2-x-1</math> would be very tough, so let's try to divide using smaller degrees of x. Doing <math>\frac{ax^3+bx^2+1}{x^2-x-1}</math>, we get the following systems of equations:
−
\begin{align*}
+
<cmath>a+b = -1,</cmath>
−
a+b & = -1\\2a+b & = 0
+
<cmath>2a+b = 0.</cmath>
−
\end{align*}
 
 
Continuing with  <math>\frac{ax^4+bx^3+1}{x^2-x-1}</math>:
 
Continuing with  <math>\frac{ax^4+bx^3+1}{x^2-x-1}</math>:
−
\begin{align*}
+
<cmath>2a+b = -1,</cmath>
−
2a+b & = -1\\3a+2b & = 0
+
<cmath>3a+2b = 0.</cmath>
−
\end{align*}
 
 
There is somewhat of a pattern showing up, so let's try <math>\frac{ax^5+bx^4+1}{x^2-x-1}</math> to verify. We get:
 
There is somewhat of a pattern showing up, so let's try <math>\frac{ax^5+bx^4+1}{x^2-x-1}</math> to verify. We get:
−
\begin{align*}
+
<cmath>3a+2b = -1,</cmath>
−
3a+2b & = -1\\5a+3b & = 0
+
<cmath>5a+3b = 0.</cmath>
−
\end{align*}
+
Now we begin to see that our pattern is actually the Fibonacci Numbers! Using the previous equations, we can make a general statement about <math>\frac{ax^n+bx^{n-1}+1}{x^2-x-1}</math>
−
Now we begin to see that our pattern is actually the Fibonacci Number's! Using the previous equations, we can make a general statement about <math>\frac{ax^n+bx^{n-1}+1}{x^2-x-1}</math>
+
<cmath>af_{n-1}+bf_{n-2} = -1,</cmath>
−
\begin{align*}
+
<cmath>af_n+bf_{n-1} = 0.</cmath>
−
af_{n-1}+bf_{n-2} & = -1\\af_n+bf_{n-1} & = 0
 
−
\end{align*}
 
 
Also, noticing our solutions from the previous systems of equations, we can create the following statement:
 
Also, noticing our solutions from the previous systems of equations, we can create the following statement:
  
−
[b]If <math>{ax^n+bx^{n-1}+1}</math> has <math>x^2-x-1</math> as a factor, then <math>a=f_{n-1}</math> and <math>b = f_n</math>[/b]
+
If <math>ax^n+bx^{n-1}+1</math> has <math>x^2-x-1</math> as a factor, then <math>a=f_{n-1}</math> and <math>b = f_n.</math>
  
−
Thus, if <math>{ax^17+bx^16+1}</math> has <math>{x^2-x-1}</math> as a factor, we get that a = 987 and b = 1597, so a = <math>\boxed {987}</math>.
+
Thus, if <math>ax^{17}+bx^{16}+1</math> has <math>x^2-x-}</math> as a factor, we get that a = 987 and b = -1597, so a = <math>\boxed {987}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:28, 29 June 2020

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 1 (Not rigorous)

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

Clearly, 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 $c_{i}$ is some 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 $a = \boxed{987}$.

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

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

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 $(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0$. 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 $a = \boxed{987}$.

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

Solution 4

The roots of $x^2-x-1$ are $\phi$ (the Golden Ratio) and $1-\phi$. These two must also be roots of $ax^{17}+bx^{16}+1$. Thus, we have two equations: $a\phi^{17}+b\phi^{16}+1=0$ and $a(1-\phi)^{17}+b(1-\phi)^{16}+1=0$. Subtract these two and divide by $\sqrt{5}$ to get $\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0$. Noting that the formula for the $n$th Fibonacci number is $\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$, we have $1597a+987b=0$. Since $1597$ and $987$ are coprime, the solutions to this equation under the integers are of the form $a=987k$ and $b=-1597k$, of which the only integral solutions for $a$ on $[0,999]$ are $0$ and $987$. $(a,b)=(0,0)$ cannot work since $x^2-x-1$ does not divide $1$, so the answer must be $\boxed{987}$. (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between $000$ and $999$).

Solution 5: For Beginners (less technical)

Trying to divide $ax^{17} + bx^{16} + 1$ by $x^2-x-1$ would be very tough, so let's try to divide using smaller degrees of x. Doing $\frac{ax^3+bx^2+1}{x^2-x-1}$, we get the following systems of equations: \[a+b = -1,\] \[2a+b = 0.\] Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$: \[2a+b = -1,\] \[3a+2b = 0.\] There is somewhat of a pattern showing up, so let's try $\frac{ax^5+bx^4+1}{x^2-x-1}$ to verify. We get: \[3a+2b = -1,\] \[5a+3b = 0.\] Now we begin to see that our pattern is actually the Fibonacci Numbers! Using the previous equations, we can make a general statement about $\frac{ax^n+bx^{n-1}+1}{x^2-x-1}$ \[af_{n-1}+bf_{n-2} = -1,\] \[af_n+bf_{n-1} = 0.\] Also, noticing our solutions from the previous systems of equations, we can create the following statement:

If $ax^n+bx^{n-1}+1$ has $x^2-x-1$ as a factor, then $a=f_{n-1}$ and $b = f_n.$

Thus, if $ax^{17}+bx^{16}+1$ has $x^2-x-}$ (Error compiling LaTeX. Unknown error_msg) as a factor, we get that a = 987 and b = -1597, so a = $\boxed {987}$.

See also

1988 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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