Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 6 (Decreases the Powers))
m (→‎Solution 8 (Engineer's Induction, only use if you don't have time left))
 
(21 intermediate revisions by 5 users not shown)
Line 63: Line 63:
 
Subtract these two and divide by <math>\sqrt{5}</math> to get <cmath>\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0.</cmath> Noting that the formula for the <math>n</math>th Fibonacci number is <math>\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}</math>, we have <math>1597a+987b=0</math>. Since <math>1597</math> and <math>987</math> are coprime, the solutions to this equation under the integers are of the form <math>a=987k</math> and <math>b=-1597k</math>, of which the only integral solutions for <math>a</math> on <math>[0,999]</math> are <math>0</math> and <math>987</math>. <math>(a,b)=(0,0)</math> cannot work since <math>x^2-x-1</math> does not divide <math>1</math>, so the answer must be <math>\boxed{987}</math>. (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between <math>000</math> and <math>999</math>).
 
Subtract these two and divide by <math>\sqrt{5}</math> to get <cmath>\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0.</cmath> Noting that the formula for the <math>n</math>th Fibonacci number is <math>\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}</math>, we have <math>1597a+987b=0</math>. Since <math>1597</math> and <math>987</math> are coprime, the solutions to this equation under the integers are of the form <math>a=987k</math> and <math>b=-1597k</math>, of which the only integral solutions for <math>a</math> on <math>[0,999]</math> are <math>0</math> and <math>987</math>. <math>(a,b)=(0,0)</math> cannot work since <math>x^2-x-1</math> does not divide <math>1</math>, so the answer must be <math>\boxed{987}</math>. (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between <math>000</math> and <math>999</math>).
  
−
== Solution 6 (Decreases the Powers) ==
+
== Solution 6 (Reduces the Powers) ==
 
We are given that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1,</math> so the roots of <math>x^2 - x - 1</math> must also be roots of <math>ax^{17} + bx^{16} + 1.</math>
 
We are given that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1,</math> so the roots of <math>x^2 - x - 1</math> must also be roots of <math>ax^{17} + bx^{16} + 1.</math>
  
−
Let <math>x=r</math> be a root of <math>x^2 - x - 1</math> so that <math>r^2 - r - 1 = 0,</math> or <math>r^2 = r + 1.</math> It follows that <cmath>ar^{17} + br^{16} + 1 = 0. \hspace{50.5mm} (\bigstar)</cmath>
+
Let <math>x=r</math> be a root of <math>x^2 - x - 1</math> so that <math>r^2 - r - 1 = 0,</math> or <math>r^2 = r + 1.</math> It follows that <cmath>ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)</cmath>
 
Note that
 
Note that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 88: Line 88:
 
987ar^2 + (610a+987b)r + 610b + 1 &= 0 \\
 
987ar^2 + (610a+987b)r + 610b + 1 &= 0 \\
 
987a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\
 
987a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\
−
(1597a+987b)r + 987a + 610b + 1 &= 0.  
+
(1597a+987b)r + (987a + 610b + 1) &= 0.  
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
Since this linear equation has two solutions of <math>r,</math> it must be an identity. Therefore, we have the following system of equations:
 
Since this linear equation has two solutions of <math>r,</math> it must be an identity. Therefore, we have the following system of equations:
Line 107: Line 107:
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 7 (Uses the Roots) ==
 +
 +
For simplicity, let <math>f(x) =ax^{17} + bx^{16} + 1</math> and <math>g(x) = x^2-x-1</math>. Notice that the roots of <math>g(x)</math> are also roots of <math>f(x)</math>. Let these roots be <math>u,v</math>. We get the system
 +
<cmath>\begin{align*}
 +
au^{17} + bu^{16} + 1 &= 0, \\
 +
av^{17} + bv^{16} + 1 &= 0.
 +
\end{align*}</cmath>
 +
If we multiply the first equation by <math>v^{16}</math> and the second by <math>u^{16}</math> we get <cmath>\begin{align*}
 +
u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\
 +
u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0.
 +
\end{align*}</cmath>
 +
Now subtracting, we get <cmath>a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \implies a = \frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.</cmath>
 +
By Vieta's, <math>uv=-1</math> so the denominator becomes <math>u-v</math>. By difference of squares and dividing out <math>u-v</math> we get <cmath>a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).</cmath> A simple exercise of Vieta's gets us <math>a= \boxed{987}.</math>
 +
 +
~bobthegod78
 +
 +
== Solution 8 (Engineer's Induction, only use if you don't have time left) ==
 +
 +
We see that <math>ax^{17} + bx^{16} + 1 = (x^2-x-1)P(x)</math> for some polynomial <math>P(x)</math>. Working forwards, we notice that the constant term of <math>P(x)</math> must equal <math>-1</math>, to multiply the constant term of <math>(x^2-x-1)</math> into <math>1</math>. We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process:
 +
<math>(x^2-x-1)(-1) = -x^2 + x + 1</math>
 +
<math>(x^2-x-1)(-1 + x) = x^3 - 2x^2 + 1</math>
 +
<math>(x^2-x-1)(-1 + x - 2x^2) = -2x^4 + 3x^3 + 1</math>
 +
<math>(x^2-x-1)(-1 + x - 2x^2 + 3x^3) = 3x^5 - 5x^4 + 1</math>.
 +
By this time, you can hopefully notice that the coefficient of the <math>x^n</math> term in <math>P(x)</math> is equal to <math>(-1)^{n-1} * F_{n}</math>, where <math>F_n</math> equals the <math>n</math>th number in the Fibonacci sequence. From here, we just need to find the coefficient of the <math>x^{15}</math> term in <math>P(x)</math>, which happens to be <math>F_{15} = \boxed{987}</math>.
 +
Again, try to only use Engineer's Induction when you have no other options. A rigorous proof is usually not needed, but when you have extra time, checking a solution with a rigorous method is better than worrying about your Engineer's Induction solution.
 +
 +
~slight edits in <math>\LaTeX</math> by [[User:Yiyj1|Yiyj1]]
  
 
== See also ==
 
== See also ==

Latest revision as of 15:30, 24 August 2024

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 (Fibonacci Numbers)

Let $F_n$ represent the $n$th number in the Fibonacci sequence. Therefore, \begin{align*} 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. \end{align*} The above uses the similarity between the Fibonacci recursion|recursive definition, $F_{n+2} - F_{n+1} - F_n = 0$, and the polynomial $x^2 - x - 1 = 0$. \begin{align*} 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 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\ &\Longrightarrow a = F_{16},\ b = - F_{17} \\ &\Longrightarrow a = \boxed {987}. \end{align*}

Solution 2 (Fibonacci Numbers)

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 polynomial with a higher degree, as shown in Solution 6.

Solution 3 (Fibonacci Numbers: 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: \begin{align*} a+b &= -1, \\ 2a+b &= 0. \end{align*} Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$: \begin{align*} 2a+b &= -1, \\ 3a+2b &= 0. \end{align*} 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: \begin{align*} 3a+2b &= -1, \\ 5a+3b &= 0. \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 $\frac{ax^n+bx^{n-1}+1}{x^2-x-1}$: \begin{align*} 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:

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 $x^2-x-1$ as a factor, we get that $a = 987$ and $b = -1597,$ so $a = \boxed {987}$.

Solution 4 (Fibonacci Numbers: 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 5 (Fibonacci Numbers)

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: \begin{align*} a\phi^{17}+b\phi^{16}+1=0, \\ a(1-\phi)^{17}+b(1-\phi)^{16}+1=0. \end{align*} 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 6 (Reduces the Powers)

We are given that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1,$ so the roots of $x^2 - x - 1$ must also be roots of $ax^{17} + bx^{16} + 1.$

Let $x=r$ be a root of $x^2 - x - 1$ so that $r^2 - r - 1 = 0,$ or $r^2 = r + 1.$ It follows that \[ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)\] Note that \begin{align*} r^4 &= (r+1)^2 \\ &= r^2 + 2r + 1 \\ &= (r+1) + 2r + 1 \\ &= 3r + 2, \\ r^8 &= (3r+2)^2 \\ &= 9r^2 + 12r + 4 \\ &= 9(r+1) + 12r + 4 \\ &= 21r + 13, \\ r^{16} &= (21r + 13)^2 \\ &= 441r^2 + 546r + 169 \\ &= 441(r+1) +546r + 169 \\ &= 987r + 610. \end{align*} We rewrite the left side of $(\bigstar)$ as a linear expression of $r:$ \begin{align*} (ar+b)r^{16} + 1 &= 0 \\ (ar+b)(987r + 610) + 1 &= 0 \\ 987ar^2 + (610a+987b)r + 610b + 1 &= 0 \\ 987a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\ (1597a+987b)r + (987a + 610b + 1) &= 0. \end{align*} Since this linear equation has two solutions of $r,$ it must be an identity. Therefore, we have the following system of equations: \begin{align*} 1597a+987b &= 0, \\ 987a+610b &= -1. \end{align*} To eliminate $b,$ we multiply the first equation by $610$ and multiply the second equation by $987,$ then subtract the resulting equations: \begin{align*} 610(1597a)+610(987b) &= 0, \\ 987(987a)+987(610b) &= -987, \end{align*} from which \begin{align*} (610\cdot1597-987\cdot987)a&=987 \\ (974170-974169)a&=987 \\ a&=\boxed{987}. \end{align*} ~MRENTHUSIASM

Solution 7 (Uses the Roots)

For simplicity, let $f(x) =ax^{17} + bx^{16} + 1$ and $g(x) = x^2-x-1$. Notice that the roots of $g(x)$ are also roots of $f(x)$. Let these roots be $u,v$. We get the system \begin{align*} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{align*} If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get \begin{align*} u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\ u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0. \end{align*} Now subtracting, we get \[a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \implies a = \frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.\] By Vieta's, $uv=-1$ so the denominator becomes $u-v$. By difference of squares and dividing out $u-v$ we get \[a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).\] A simple exercise of Vieta's gets us $a= \boxed{987}.$

~bobthegod78

Solution 8 (Engineer's Induction, only use if you don't have time left)

We see that $ax^{17} + bx^{16} + 1 = (x^2-x-1)P(x)$ for some polynomial $P(x)$. Working forwards, we notice that the constant term of $P(x)$ must equal $-1$, to multiply the constant term of $(x^2-x-1)$ into $1$. We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process: $(x^2-x-1)(-1) = -x^2 + x + 1$ $(x^2-x-1)(-1 + x) = x^3 - 2x^2 + 1$ $(x^2-x-1)(-1 + x - 2x^2) = -2x^4 + 3x^3 + 1$ $(x^2-x-1)(-1 + x - 2x^2 + 3x^3) = 3x^5 - 5x^4 + 1$. By this time, you can hopefully notice that the coefficient of the $x^n$ term in $P(x)$ is equal to $(-1)^{n-1} * F_{n}$, where $F_n$ equals the $n$th number in the Fibonacci sequence. From here, we just need to find the coefficient of the $x^{15}$ term in $P(x)$, which happens to be $F_{15} = \boxed{987}$. Again, try to only use Engineer's Induction when you have no other options. A rigorous proof is usually not needed, but when you have extra time, checking a solution with a rigorous method is better than worrying about your Engineer's Induction solution.

~slight edits in $\LaTeX$ by Yiyj1

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_13&oldid=225690"