Difference between revisions of "1988 AIME Problems/Problem 13"
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− | + | Use [[Indeterminate Coefficients]]: 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>. | |
− | + | 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_{15}</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>. | |
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 = 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 = 987</math>. |
Revision as of 12:26, 2 July 2016
Problem
Find if and are integers such that is a factor of .
Solution 1
Use Indeterminate Coefficients: 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 ).
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.