Difference between revisions of "1988 AIME Problems/Problem 13"
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== Problem == | == 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 [[integer]]s such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>. | ||
− | + | == Solution 1 (Not rigorous)== | |
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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>. | 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>. | ||
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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>. | 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>. | ||
− | + | == Solution 2 == | |
Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore, | Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore, | ||
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<math>a = F_{16},\ b = - F_{17}\Longrightarrow a = \boxed {987}\ .</math> | <math>a = F_{16},\ b = - F_{17}\Longrightarrow a = \boxed {987}\ .</math> | ||
− | + | == Solution 3 == | |
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}b + 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 = \boxed{987}</math>. | 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}b + 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 = \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 larger polynomial. | 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. | ||
− | + | == 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: | 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: | ||
<cmath>a+b = -1,</cmath> | <cmath>a+b = -1,</cmath> |
Latest revision as of 20:22, 17 April 2021
Contents
Problem
Find if and are integers such that is a factor of .
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: 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.