Difference between revisions of "1988 AIME Problems/Problem 13"
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.) |
MRENTHUSIASM (talk | contribs) m (→Solution 3 (Fibonacci Version 3: For Beginners, Less Technical)) |
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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> | 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> | ||
− | 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 | + | 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 <math>a = \boxed {987}</math>. |
== Solution 4 (Fibonacci Version 4: Not Rigorous)== | == Solution 4 (Fibonacci Version 4: Not Rigorous)== |
Revision as of 10:30, 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 ).
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.