2009 AMC 12B Problems/Problem 19
Contents
Problem
For each positive integer , let
. What is the sum of all values of
that are prime numbers?
Solution
Solution 1
To find the answer it was enough to play around with . One can easily find that
is a prime, then
becomes negative for
between
and
, and then
is again a prime number. And as
is already the largest option, the answer must be
.
Solution 2
We will now show a complete solution, with a proof that no other values are prime.
Consider the function , then obviously
.
The roots of are:
We can then write , and thus
.
We would now like to factor the right hand side further, using the formula . To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.
We are looking for rational and
such that
. Expanding the left hand side and comparing coefficients, we get
and
. We can easily guess (or compute) the solution
,
.
Hence , and we can easily verify that also
.
We now know the complete factorization of :
As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.
We have ,
and
.
Hence we obtain the factorization .
For both terms are positive and larger than one, hence
is not prime. For
the second factor is positive and the first one is negative, hence
is not a prime. The remaining cases are
and
. In both cases,
is indeed a prime, and their sum is
.
Solution 3
Instead of doing the hard work, we can try to guess the factorization. One good approach:
We can make the observation that looks similar to
with the exception of the
term. In fact, we have
. But then we notice that it differs from the desired expression by a square:
.
Now we can use the formula to obtain the same factorization as in the previous solution, without all the work.
Solution 4
After arriving at the factorization , a more mathematical approach would be to realize that the second factor is always positive when
is a positive integer. Therefore, in order for
to be prime, the first factor has to be
.
We can set it equal to 1 and solve for :
Substituting these values into the second factor and adding would give the answer.
Solution 5 (Using the answer choices)
Plugging in a few values, it is obvious that is prime. Starting from
, we know that
becomes negative, then positive again. We want the find the first integral
in which
is positive. This is solved by
.
Instead of using the quadratic formula, we can ignore to get an approximation. So, we want to solve
.
Simplifying,
. Looking at this, we can automatically see that
is the first value that this happens.
We are only looking for odd values of . Although we can check if
is positive to account for the
we took off, this will be unnecessary because it wouldn’t be prime anyways.
Plugging in , we get
, which is prime.
We can establish that is prime in two ways. Firstly, we can check prime values until
to see if this is true.
The second way is faster, but requires the answer choices. We know that and
is the largest answer choice. Since
is the first odd value of
greater than 0, it follows that
has to be prime for the answer choices to work.
As a result, the answer is .
See Also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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