Difference between revisions of "1988 AIME Problems/Problem 13"
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(Another more easier solution for beginners to AIME problems, utilizing the fibonacci pattern.) |
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=== Solution 4 === | === Solution 4 === | ||
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>). | 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>). | ||
+ | |||
+ | === 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: | ||
+ | \begin{align*} | ||
+ | a+b & = -1\\2a+b & = 0 | ||
+ | \end{align*} | ||
+ | Continuing with <math>\frac{ax^4+bx^3+1}{x^2-x-1}</math>: | ||
+ | \begin{align*} | ||
+ | 2a+b & = -1\\3a+2b & = 0 | ||
+ | \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: | ||
+ | \begin{align*} | ||
+ | 3a+2b & = -1\\5a+3b & = 0 | ||
+ | \end{align*} | ||
+ | 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> | ||
+ | \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: | ||
+ | |||
+ | [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] | ||
+ | |||
+ | 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>. | ||
== See also == | == See also == |
Revision as of 13:29, 22 June 2020
Problem
Find if and are integers such that is a factor of .
Contents
Solution 1 (Not rigorous)
Let's work backwards! Let and let be the polynomial such that .
Clearly, the constant term of must be . Now, we have , where is some coefficient. However, since has no term, it must be true that .
Let's find now. Notice that all we care about in finding is that . Therefore, . Undergoing a similar process, , , , and we see a nice pattern. The coefficients of are just the Fibonacci sequence with alternating signs! Therefore, , where denotes the 16th Fibonnaci number and .
Solution 2
Let represent the th number in the Fibonacci sequence. Therefore,
The above uses the similarity between the Fibonacci recursive definition, , and the polynomial .
and
Solution 3
We can long divide and search for a pattern; then the remainder would be set to zero to solve for . 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 . Since the coefficient of must be zero, this gives us two equations, and . Solving these two as above, we get that .
There are various similar solutions which yield the same pattern, such as repeated substitution of into the larger polynomial.
Solution 4
The roots of are (the Golden Ratio) and . These two must also be roots of . Thus, we have two equations: and . Subtract these two and divide by to get . Noting that the formula for the th Fibonacci number is , we have . Since and are coprime, the solutions to this equation under the integers are of the form and , of which the only integral solutions for on are and . cannot work since does not divide , so the answer must be . (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between and ).
Solution 5: For Beginners (less technical)
Trying to divide by would be very tough, so let's try to divide using smaller degrees of x. Doing , we get the following systems of equations: \begin{align*} a+b & = -1\\2a+b & = 0 \end{align*} Continuing with : \begin{align*} 2a+b & = -1\\3a+2b & = 0 \end{align*} There is somewhat of a pattern showing up, so let's try 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 Number's! Using the previous equations, we can make a general statement about \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:
[b]If has as a factor, then and [/b]
Thus, if has as a factor, we get that a = 987 and b = 1597, so a = .
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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