Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is | To complete the solution, we can let <math>a = 315</math>, and then try to find <math>Q(x)</math>. We know from the above calculation that <math>Q(2)=42, Q(4)=-70, Q(6)=42</math>, and <math>Q(8)=-6</math>. Then we can let <math>Q(x) = T(x)(x-2)(x-6)+42</math>, getting <math>T(4)=28, T(8)=-4</math>. Let <math>T(x)=L(x)(x-8)-4</math>, then <math>L(4)=-8</math>. Therefore, it is possible to choose <math>T(x) = -8(x-8)-4 = -8x + 60</math>, so the goal is accomplished. As a reference, the polynomial we get is | ||
− | <cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315 = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325</cmath> | + | <cmath>P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315</cmath> |
+ | <cmath> = -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325</cmath> | ||
== See also == | == See also == |
Revision as of 23:46, 30 December 2015
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 the Critique
Once you find that is a factor of 315, you can instantly find that the answer is 315 because the problem asks for the smallest value of
.
Critique of the (Critique of the 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.