Difference between revisions of "1988 AIME Problems/Problem 13"
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== Solution 3 (Fibonacci Version 3: For Beginners, Less Technical) == | == Solution 3 (Fibonacci Version 3: 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 <math>x</math>. 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>\begin{align*} |
− | + | a+b &= -1, \ | |
− | Continuing with | + | 2a+b &= 0. |
− | <cmath>2a+b = -1, | + | \end{align*}</cmath> |
− | + | Continuing with <math>\frac{ax^4+bx^3+1}{x^2-x-1}</math>: | |
+ | <cmath>\begin{align*} | ||
+ | 2a+b &= -1, \ | ||
+ | 3a+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: | 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>3a+2b = -1, | + | <cmath>\begin{align*} |
− | + | 3a+2b &= -1, \ | |
− | 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> | + | 5a+3b &= 0. |
− | <cmath>af_{n-1}+bf_{n-2} = -1, | + | \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: | Also, noticing our solutions from the previous systems of equations, we can create the following statement: | ||
Revision as of 11:01, 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 . 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.