Difference between revisions of "1988 AIME Problems/Problem 13"
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== Solution 5 (Fibonacci Version 5) == | == Solution 5 (Fibonacci Version 5) == | ||
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 6 (Decreases 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> are roots of <math>ax^{17} + bx^{16} + 1.</math> | ||
+ | |||
+ | Let <math>x=t</math> be a root of <math>x^2 - x - 1</math> so that <math>t^2 - t - 1 = 0,</math> or <math>t^2 = t + 1.</math> Note that | ||
+ | <cmath>\begin{align*} | ||
+ | t^4 &= (t+1)^2 \ | ||
+ | &= t^2 + 2t + 1 \ | ||
+ | &= (t+1) + 2t + 1 \ | ||
+ | &= 3t + 2, \ | ||
+ | t^8 &= (3t+2)^2 \ | ||
+ | &= 9t^2 + 12t + 4 \ | ||
+ | &= 9(t+1) + 12t + 4 \ | ||
+ | &= 21t + 13, \ | ||
+ | t^{16} &= (21t + 13)^2 \ | ||
+ | &= 441t^2 + 546t + 169 \ | ||
+ | &= 441(t+1) +546t + 169 \ | ||
+ | &= 987t + 610. | ||
+ | \end{align*}</cmath> | ||
== See also == | == See also == |
Revision as of 10:56, 30 June 2021
Contents
[hide]Problem
Find if and are integers such that is a factor of .
Solution 1 (Fibonacci Version 1)
Let represent the th number in the Fibonacci sequence. Therefore,
The above uses the similarity between the Fibonacci recursion|recursive definition, , and the polynomial .
and
Solution 2 (Fibonacci Version 2)
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 3 (Fibonacci Version 3: 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: Continuing with : There is somewhat of a pattern showing up, so let's try to verify. We get: Now we begin to see that our pattern is actually the Fibonacci Numbers! Using the previous equations, we can make a general statement about Also, noticing our solutions from the previous systems of equations, we can create the following statement:
If has as a factor, then and
Thus, if has as a factor, we get that and so .
Solution 4 (Fibonacci Version 4: 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 5 (Fibonacci Version 5)
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 6 (Decreases the Powers)
We are given that is a factor of so the roots of are roots of
Let be a root of so that or Note that
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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