Difference between revisions of "1977 AHSME Problems/Problem 28"
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<cmath>g(x^{12}) = 6</cmath> | <cmath>g(x^{12}) = 6</cmath> | ||
<cmath>g(x) = 0</cmath> | <cmath>g(x) = 0</cmath> | ||
− | We have <math>5</math> values of <math>x</math> that return <math>R(x) = 6</math>. However, <math>g(x)</math> is quintic, implying the remainder is of degree <math>4</math> — contradicted by the <math>5</math> solutions. Thus, the only remaining possibility is that the remainder is a constant <math>\boxed{ | + | We have <math>5</math> values of <math>x</math> that return <math>R(x) = 6</math>. However, <math>g(x)</math> is quintic, implying the remainder is of degree <math>4</math> — contradicted by the <math>5</math> solutions. Thus, the only remaining possibility is that the remainder is a constant <math>\boxed{6}</math>. |
Revision as of 20:42, 5 August 2019
Solution 1
Let be the remainder when
is divided by
. Then
is the unique polynomial such that
is divisible by
, and
.
Note that is a multiple of
. Also,
Each term is a multiple of
. For example,
Hence,
is a multiple of
, which means that
is a multiple of
. Therefore, the remainder is
. The answer is (A).
Solution 2
We express the quotient and remainder as follows.
Note that the solutions to
correspond to the 6th roots of unity, excluding
. Hence, we have
, allowing us to set:
We have
values of
that return
. However,
is quintic, implying the remainder is of degree
— contradicted by the
solutions. Thus, the only remaining possibility is that the remainder is a constant
.