Difference between revisions of "2009 AMC 12B Problems/Problem 19"
IMOJonathan (talk | contribs) (→Solution 5 (Using the answer choices)) |
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− | Plugging in <math>n = 19</math>, we get <math>f(19) = 19^2(19^2-360)+400 = 19^2*1+400 = 761</math>. Since we know that <math>41+761 = 802</math> and <math>802</math> is the largest answer choice, it follows that the answer is <math> | + | Plugging in <math>n = 19</math>, we get <math>f(19) = 19^2(19^2-360)+400 = 19^2*1+400 = 761</math>. Since we know that <math>41+761 = 802</math> and <math>802</math> is the largest answer choice, it follows that the answer is <math>\boxed{802}</math>. |
== See Also == | == See Also == |
Revision as of 14:31, 15 July 2015
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
. Since we know that
and
is the largest answer choice, it follows that 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.