Difference between revisions of "1988 AIME Problems/Problem 13"
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MRENTHUSIASM (talk | contribs) (Rearranged the solutions based on rigor and education values. I also deleted the links directed to the other pages, as they are not necessary. If anyone is not happy with this result, please PM me.) |
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== Problem == | == Problem == | ||
− | Find <math>a</math> if <math>a</math> and <math>b</math> are | + | 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>. |
− | == Solution 1 ( | + | == Solution 1 (Fibonacci Version 1) == |
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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, | ||
<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> | <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> | ||
− | The above uses the similarity between the Fibonacci | + | 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>. |
<math>0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow</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> | ||
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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 | + | == 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 <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 | + | == Solution 3 (Fibonacci Version 3: For Beginners, Less Technical) == |
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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> | ||
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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 a = <math>\boxed {987}</math>. | 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 a = <math>\boxed {987}</math>. | ||
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+ | == Solution 4 (Fibonacci Version 4: 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>. | ||
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+ | 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_{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>. | ||
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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>. | ||
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+ | == 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>). | ||
== See also == | == See also == |
Revision as of 11:26, 30 June 2021
Contents
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 a =
.
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
).
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.