Difference between revisions of "2010 AMC 10B Problems/Problem 25"
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+ | == Problem == | ||
+ | Let <math>a > 0</math>, and let <math>P(x)</math> be a polynomial with integer coefficients such that | ||
+ | |||
+ | <center> | ||
+ | <math>P(1) = P(3) = P(5) = P(7) = a</math>, and<br/> | ||
+ | <math>P(2) = P(4) = P(6) = P(8) = -a</math>. | ||
+ | </center> | ||
+ | |||
+ | What is the smallest possible value of <math>a</math>? | ||
+ | |||
+ | <math>\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!</math> | ||
+ | |||
+ | == Solution == | ||
There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math> | There must be some polynomial <math>Q(x)</math> such that <math>P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x)</math> | ||
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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>. | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2010|ab=B|num-b=24|after=Last question}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 23:23, 21 April 2010
Problem
Let , and let
be a polynomial with integer coefficients such that
, and
.
What is the smallest possible value of ?
Solution
There must be some 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
.
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 |