Difference between revisions of "1988 AIME Problems/Problem 13"
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− | We can long divide and search for a pattern; then the remainder would be set to zero to solve for <math>a</math>. 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 <math>(F_{16} + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0</math>. Since the coefficient of <math>x</math> must be zero, this gives us two equations, <math>F_{16}b + F_{17}a = 0</math> and <math>F_{15}b + F_{16}a + 1 = 0</math>. Solving these two as above, we get that <math>a = 987</math>. | + | We can long divide and search for a pattern; then the remainder would be set to zero to solve for <math>a</math>. 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 <math>(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0</math>. Since the coefficient of <math>x</math> must be zero, this gives us two equations, <math>F_{16}b + F_{17}a = 0</math> and <math>F_{15}b + F_{16}a + 1 = 0</math>. Solving these two as above, we get that <math>a = 987</math>. |
There are various similar solutions which yield the same pattern, such as repeated substitution of <math>x^2 = x + 1</math> into the larger polynomial. | There are various similar solutions which yield the same pattern, such as repeated substitution of <math>x^2 = x + 1</math> into the larger polynomial. |
Revision as of 19:28, 16 March 2017
Problem
Find if
and
are integers such that
is a factor of
.
Solution 1
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 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.