Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>. | Thus, the least value of <math>a</math> must be the <math>lcm(15,9,15,105)</math>. | ||
Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. | Solving, we receive <math>315</math>, so our answer is <math> \boxed{\textbf{(B)}\ 315} </math>. | ||
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== Critique of Critique== | == Critique of Critique== |
Revision as of 21:13, 21 December 2016
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 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
See also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 10 Problems and Solutions |
See also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.