1988 AIME Problems/Problem 13
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, 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
. But the formula for the nth fibonacci number is
(You may want to research this). Thus, we have
, so since
and
are relatively prime, and the anwser must be a positive integer less than
, we can guess that it equals
.
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 | ||
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