2010 AMC 10B Problems/Problem 25
Contents
Problem
Let , and let
be a polynomial with integer coefficients such that
, and
.
What is the smallest possible value of ?
Solution
We observe that because , if we define a new polynomial
such that
,
has roots when
; namely, when
.
Thus since has roots when
, we can factor the product
out of
to obtain a new polynomial
such that
.
Then, plugging in values of we get
Thus, the least value of
must be the
.
Solving, we receive
, so our answer is
.
Critique
The above solution is incomplete. What is really proven is that 315 is a factor of , if such an
exists. That only rules out answer A.
To prove that the answer is correct, one could exhibit a polynomium that satisfies the requirements with . Here's one:
.
You get that from the matrix
and
and computing
which comes out as the all-integer coefficients above.
Critique of Critique
First of all, the solution shows that is a multiple of
, not a factor of
. Many people confuse the usage of the words 'factor' and 'multiple'. Secondly, even if
is a multiple of
, it does not mean that you can instantly get that the answer is
because we need to know that
is possible. After all,
is also a multiple of
, but
is definitely not the smallest possible number.
To complete the solution, we can let , and then try to find
. We know from the above calculation that
, and
. Then we can let
, getting
. Let
, then
. Therefore, it is possible to choose
, so the goal is accomplished. As a reference, the polynomial we get is
Critique of Critique of Critique
The problem states the "least value" of , so it is not needed to add the extra steps.
Critique of Critique of Critique of Critique
It is still necessary to show that the minimum is achievable.
For example , but
is not the least value of
Critique of all the previous Critiques
Why are there so many?
See also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
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Followed by Last question | |
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All AMC 10 Problems and Solutions |
See also
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