1988 AIME Problems/Problem 13
Contents
- 1 Problem
- 2 Solution 1 (Fibonacci Numbers)
- 3 Solution 2 (Fibonacci Numbers)
- 4 Solution 3 (Fibonacci Numbers: For Beginners, Less Technical)
- 5 Solution 4 (Fibonacci Numbers: Not Rigorous)
- 6 Solution 5 (Fibonacci Numbers)
- 7 Solution 6 (Reduces the Powers)
- 8 Solution 7 (Uses the Roots)
- 9 Solution 8 (Engineer's Induction, only use if you don't have time left)
- 10 See also
Problem
Find if and are integers such that is a factor of .
Solution 1 (Fibonacci Numbers)
Let represent the th number in the Fibonacci sequence. Therefore, The above uses the similarity between the Fibonacci recursion|recursive definition, , and the polynomial .
Solution 2 (Fibonacci Numbers)
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 polynomial with a higher degree, as shown in Solution 6.
Solution 3 (Fibonacci Numbers: 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 Numbers: 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 Numbers)
The roots of are (the Golden Ratio) and . These two must also be roots of . Thus, we have two equations: 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 (Reduces the Powers)
We are given that is a factor of so the roots of must also be roots of
Let be a root of so that or It follows that Note that We rewrite the left side of as a linear expression of Since this linear equation has two solutions of it must be an identity. Therefore, we have the following system of equations: To eliminate we multiply the first equation by and multiply the second equation by then subtract the resulting equations: from which ~MRENTHUSIASM
Solution 7 (Uses the Roots)
For simplicity, let and . Notice that the roots of are also roots of . Let these roots be . We get the system If we multiply the first equation by and the second by we get Now subtracting, we get By Vieta's, so the denominator becomes . By difference of squares and dividing out we get A simple exercise of Vieta's gets us
~bobthegod78
Solution 8 (Engineer's Induction, only use if you don't have time left)
We see that for some polynomial . Working forwards, we notice that the constant term of must equal , to multiply the constant term of into . We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process: . By this time, you can hopefully notice that the coefficient of the term in is equal to , where equals the th number in the Fibonacci sequence. From here, we just need to find the coefficient of the term in , which happens to be . Again, try to only use Engineer's Induction when you have no other options. A rigorous proof is usually not needed, but when you have extra time, checking a solution with a rigorous method is better than worrying about your Engineer's Induction solution.
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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