Difference between revisions of "1988 AIME Problems/Problem 13"
Dgreenb801 (talk | contribs) (→Solution) |
Dgreenb801 (talk | contribs) m (→Solution 4) |
||
Line 32: | Line 32: | ||
=== Solution 4 === | === Solution 4 === | ||
− | The roots of <math>x^2-x-1</math> are the golden ratio and 1-(the golden ratio). Call these <math>g</math> and <math>1-g</math>. These two must also be roots of <math>ax^{17}+bx^{16}+1</math>. Thus, we have two equations: <math>ag^{17}+bg^{16}+1=0</math> and <math>a(1-g)^{17}+b(1-g)^{16}+1=0</math>. Subtract these two and divide by <math>\sqrt{5}</math> to get <math>a(g^{17}-(1-g)^{17})/\sqrt{5}+b(g^{16}-(1-g)^{16})/\sqrt{5}=0</math>. But the formula for the nth fibonacci number is <math>(g^n-(1-g)^n)/\sqrt{5})</math> (You may want to research this). Thus, we have <math>1597a+987b=0</math>, so since | + | The roots of <math>x^2-x-1</math> are the golden ratio and 1-(the golden ratio). Call these <math>g</math> and <math>1-g</math>. These two must also be roots of <math>ax^{17}+bx^{16}+1</math>. Thus, we have two equations: <math>ag^{17}+bg^{16}+1=0</math> and <math>a(1-g)^{17}+b(1-g)^{16}+1=0</math>. Subtract these two and divide by <math>\sqrt{5}</math> to get <math>a(g^{17}-(1-g)^{17})/\sqrt{5}+b(g^{16}-(1-g)^{16})/\sqrt{5}=0</math>. But the formula for the nth fibonacci number is <math>(g^n-(1-g)^n)/\sqrt{5})</math> (You may want to research this). Thus, we have <math>1597a+987b=0</math>, so since 1597 and 987 are relatively prime, and the anwser must be a positive integer less than 1000, we can guess it equals 987. |
== See also == | == See also == |
Revision as of 07:13, 3 April 2008
Problem
Find if and are integers such that is a factor of .
Contents
Solution
Solution 1
Let's work backwards! Let and let be the polynomial such that .
First, it's kinda obvious that the constant term of must be . Now, we have , where is some random 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 1-(the golden ratio). Call these and . These two must also be roots of . Thus, we have two equations: and . Subtract these two and divide by to get . But the formula for the nth fibonacci number is (You may want to research this). Thus, we have , so since 1597 and 987 are relatively prime, and the anwser must be a positive integer less than 1000, we can guess it equals 987.
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 |