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

Latest revision as of 15:30, 24 August 2024

1 Problem
2 Solution 1 (Fibonacci Numbers)
3 Solution 2 (Fibonacci Numbers)
4 Solution 3 (Fibonacci Numbers: For Beginners, Less Technical)
5 Solution 4 (Fibonacci Numbers: Not Rigorous)
6 Solution 5 (Fibonacci Numbers)
7 Solution 6 (Reduces the Powers)
8 Solution 7 (Uses the Roots)
9 Solution 8 (Engineer's Induction, only use if you don't have time left)
10 See also

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

@@ 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>.
+Find <math>a</math> if <math>a</math> and <math>b</math> are integers such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>.
-__TOC__
+== Solution 1 (Fibonacci Numbers) ==
-=== Solution 1 ===
+Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore,
-Use [[Indeterminate Coefficients]]: 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>.
+<cmath>\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*}</cmath>
+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>.
+<cmath>\begin{align*}
+= 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*}</cmath>
+== Solution 2 (Fibonacci Numbers) ==
+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 <cmath>(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.</cmath> 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 = \boxed{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 polynomial with a higher degree, as shown in Solution 6.
-Clearly, 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 [[coefficient]]. However, since <math>F(x)</math> has no <math>x</math> term, it must be true that <math>c_{15} = 1</math>.
+== Solution 3 (Fibonacci Numbers: 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 <math>x</math>. Doing <math>\frac{ax^3+bx^2+1}{x^2-x-1}</math>, we get the following systems of equations:
+<cmath>\begin{align*}
+a+b &= -1, \\
+a+b &= 0.
+\end{align*}</cmath>
+Continuing with <math>\frac{ax^4+bx^3+1}{x^2-x-1}</math>:
+<cmath>\begin{align*}
+a+b &= -1, \\
+a+2b &= 0.
+\end{align*}</cmath>
+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:
+<cmath>\begin{align*}
+a+2b &= -1, \\
+a+3b &= 0.
+\end{align*}</cmath>
+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>:
+<cmath>\begin{align*}
+af_{n-1}+bf_{n-2} &= -1, \\
+af_n+bf_{n-1} &= 0.
+\end{align*}</cmath>
+Also, noticing our solutions from the previous systems of equations, we can create the following statement:
-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>.
+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>
-=== Solution 2 ===
+Thus, if <math>ax^{17}+bx^{16}+1</math> has <math>x^2-x-1</math> as a factor, we get that <math>a = 987</math> and <math>b = -1597,</math> so <math>a = \boxed {987}</math>.
-Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore,
+== Solution 4 (Fibonacci Numbers: Not Rigorous)==
+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>.
+Clearly, the constant term of <math>P(x)</math> must be <math>- 1</math>. Now, we have <cmath>(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1),</cmath> where <math>c_{i}</math> is some 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 = \boxed{987}</math>.
-<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>
+== Solution 5 (Fibonacci Numbers) ==
+The roots of <math>x^2-x-1</math> are <math>\phi</math> (the Golden Ratio) and <math>1-\phi</math>. These two must also be roots of <math>ax^{17}+bx^{16}+1</math>. Thus, we have two equations:
+<cmath>\begin{align*}
+a\phi^{17}+b\phi^{16}+1=0, \\
+a(1-\phi)^{17}+b(1-\phi)^{16}+1=0.
+\end{align*}</cmath>
+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>).
-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>.
+== 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>
-<math>0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow</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{20mm} (\bigstar)</cmath>
+Note that
+<cmath>\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*}</cmath>
+We rewrite the left side of <math>(\bigstar)</math> as a linear expression of <math>r:</math>
+<cmath>\begin{align*}
+(ar+b)r^{16} + 1 &= 0 \\
+(ar+b)(987r + 610) + 1 &= 0 \\
+ar^2 + (610a+987b)r + 610b + 1 &= 0 \\
+a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\
+(1597a+987b)r + (987a + 610b + 1) &= 0.
+\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:
+<cmath>\begin{align*}
+a+987b &= 0, \\
+a+610b &= -1.
+\end{align*}</cmath>
+To eliminate <math>b,</math> we multiply the first equation by <math>610</math> and multiply the second equation by <math>987,</math> then subtract the resulting equations:
+<cmath>\begin{align*}
+(1597a)+610(987b) &= 0, \\
+(987a)+987(610b) &= -987,
+\end{align*}</cmath>
+from which
+<cmath>\begin{align*}
+(610\cdot1597-987\cdot987)a&=987 \\
+(974170-974169)a&=987 \\
+a&=\boxed{987}.
+\end{align*}</cmath>
+~MRENTHUSIASM
-<math>(aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q\Longrightarrow</math>
+== Solution 7 (Uses the Roots) ==
-<math>aF_{17} + bF_{16} = 0</math> and <math>aF_{16} + bF_{15} + 1 = 0\Longrightarrow</math>
+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>
-<math>a = F_{16},\ b = - F_{17}\Longrightarrow \boxed {a = 987}\ .</math>
+~bobthegod78
-=== Solution 3 ===
+== Solution 8 (Engineer's Induction, only use if you don't have time left) ==
-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.
+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.
-=== Solution 4 ===
+~slight edits in <math>\LaTeX</math> by [[User:Yiyj1|Yiyj1]]
-The roots of <math>x^2-x-1</math> are <math>\phi</math> (the [[Golden Ratio]]) and <math>1-\phi</math>. These two must also be roots of <math>ax^{17}+bx^{16}+1</math>. Thus, we have two equations: <math>a\phi^{17}+b\phi^{16}+1=0</math> and <math>a(1-\phi)^{17}+b(1-\phi)^{16}+1=0</math>. Subtract these two and divide by <math>\sqrt{5}</math> to get <math>\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0</math>. 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>).
 == 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

Page

Toolbox

Search