2016 AIME I Problems/Problem 11
Contents
Problem
Let be a nonzero polynomial such that
for every real
, and
. Then
, where
and
are relatively prime positive integers. Find
.
Solution 1
Plug in to get
. Plug in
to get
. Plug in
to get
. So
for some polynomial
. Using the initial equation, once again,
From here, we know that
for a constant
(
cannot be periodic since it is a polynomial), so
. We know that
. Plugging those into our definition of
:
or
. So we know that
. So
. Thus, the answer is
.
Solution 2
From the equation we see that divides
and
divides
so we can conclude that
and
divide
(if we shift the function right by 1, we get
, and from here we can see that
divides
). This means that
and
are roots of
. Plug in
and we see that
so
is also a root.
Suppose we had another root that is not one of those . Notice that the equation above indicates that if
is a root then
and
is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.
That means . We can use
to get
. Plugging in
is now trivial and we see that it is
so our answer is
Solution 3
Although this may not be the most mathematically rigorous answer, we see that . Using a bit of logic, we can make a guess that
has a factor of
, telling us
has a factor of
. Similarly, we guess that
has a factor of
, which means
has a factor of
. Now, since
and
have so many factors that are off by one, we may surmise that when you plug
into
, the factors "shift over," i.e.
, which goes to
. This is useful because these, when divided, result in
. If
, then we get
and
,
. This gives us
and
, and at this point we realize that there has to be some constant
multiplied in front of the factors, which won't affect our fraction
but will give us the correct values of
and
. Thus
, and we utilize
to find
. Evaluating
is then easy, and we see it equals
, so the answer is
Solution 4
Substituting into the given equation, we find that
. Therefore, either
or
. Now for integers
, we know that
Applying this repeatedly, we find that
If
, this shows that
has infinitely many roots, meaning that
is identically equal to zero. But this contradicts the problem statement. Therefore,
, and we find
for all positive integers
. This cubic polynomial matches the values
for infinitely many numbers, hence the two polynomials are identically equal. In particular,
, and the answer is
.
Solution 5
We can find zeroes of the polynomial by making the first given equation . Plugging in
and
gives us the zeroes
and
, respectively. Now we can plug in these zeros to get more zeroes.
gives us the zero
(no pun intended).
makes the equation
, which means
is not necessarily
. If
, then plugging in
to the equation yields
, plugging in
to the equation yields
, and so on, a contradiction of "nonzero polynomial". So
is not a zero. Note that plugging in
to the equation does not yield any additional zeros. Thus, the only zeroes of
are
and
, so
for some nonzero constant
. We can plug in
and
into the polynomial and use the second given equation to find an equation for
.
and
, so:
Plugging in
into the polynomial
yields
.
.
Solution 6 (very Easy)
Plug in yields
. Since also
, we have
and
. Plug in
yields
so
.
Repeat the action gives ,
,
,
, and
.
Since is a polynomial, the
th difference is constant, where
. Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant.
Since the 3rd difference of is constant, we can conclude that
.
Let . Plug in the values for
and solve the system of 4 equations gives
Thus and $P(\frac{7}{2})=\frac{105}{4}\Longrightarrow m+n=\boxed{109}
~ Nafer
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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