Difference between revisions of "Mock AIME I 2012 Problems/Problem 12"
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==Solution== | ==Solution== | ||
Notice that if <math>a</math> is a root of <math>P</math>, then <math>a^2</math> must be a root of <math>P</math> and <math>(a + 1)^2</math> must be a root of <math>P</math>. But then continuing this, <math>a^{2^n}</math> and <math>(a + 1)^{2^n}</math> must be roots of <math>P</math> for all <math>n</math>. Since a polynomial has finitely many roots, <math>a</math> and <math>a + 1</math> must be roots of unity so that the above two sets contain finitely many elements. But there is a unique such pair of roots of unity, making <math>a = </math> <math>-1/2 \pm i\sqrt{3}/2</math>. Then the disjoint union of the two sets above is <math>\{-1/2 + i\sqrt{3}/2, 1/2 + i\sqrt{3}/2\}</math>, the minimal polynomial for which is <math>x^2 + x + 1</math>. Since any power of this base polynomial will work, <math>P(x) = (x^2 + x + 1)^5</math>, making the sum of coefficients <math>\boxed{243}</math>. | Notice that if <math>a</math> is a root of <math>P</math>, then <math>a^2</math> must be a root of <math>P</math> and <math>(a + 1)^2</math> must be a root of <math>P</math>. But then continuing this, <math>a^{2^n}</math> and <math>(a + 1)^{2^n}</math> must be roots of <math>P</math> for all <math>n</math>. Since a polynomial has finitely many roots, <math>a</math> and <math>a + 1</math> must be roots of unity so that the above two sets contain finitely many elements. But there is a unique such pair of roots of unity, making <math>a = </math> <math>-1/2 \pm i\sqrt{3}/2</math>. Then the disjoint union of the two sets above is <math>\{-1/2 + i\sqrt{3}/2, 1/2 + i\sqrt{3}/2\}</math>, the minimal polynomial for which is <math>x^2 + x + 1</math>. Since any power of this base polynomial will work, <math>P(x) = (x^2 + x + 1)^5</math>, making the sum of coefficients <math>\boxed{243}</math>. | ||
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+ | ==See Also== | ||
+ | *[[Mock AIME I 2012 Problems/Problem 13| Next Problem]] | ||
+ | *[[Mock AIME I 2012 Problems/Problem 11| Previous Problem]] | ||
+ | *[[Mock AIME I 2012 Problems]] |
Latest revision as of 10:00, 11 March 2025
Problem
Let be a polynomial of degree 10 satisfying
. Find the maximum possible sum of the coefficients of
.
Solution
Notice that if is a root of
, then
must be a root of
and
must be a root of
. But then continuing this,
and
must be roots of
for all
. Since a polynomial has finitely many roots,
and
must be roots of unity so that the above two sets contain finitely many elements. But there is a unique such pair of roots of unity, making
. Then the disjoint union of the two sets above is
, the minimal polynomial for which is
. Since any power of this base polynomial will work,
, making the sum of coefficients
.