Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 13"
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It becomes obvious that <math>F(x) = ax(x-1)</math>, for some [[constant]] <math>a</math>, matches the definition of the polynomial. To prove that <math>F(x)</math> must have this form, note that | It becomes obvious that <math>F(x) = ax(x-1)</math>, for some [[constant]] <math>a</math>, matches the definition of the polynomial. To prove that <math>F(x)</math> must have this form, note that | ||
<cmath>(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)</cmath> | <cmath>(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)</cmath> | ||
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+ | Since <math>3x</math> and <math>3x-1</math> divides the right side of the equation, <math>3x</math> and <math>3x-1</math> divides the left side of the equation. Thus <math>3x(3x-1)</math> divides <math>F(3x)</math>, so <math>x(x-1)</math> divides <math>F(x)</math>. | ||
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+ | It is easy to see that <math>F(x)</math> is a [[quadratic]], thus <math>F(x)=ax(x-1)</math> as desired. | ||
By the given, <math>F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12</math>. Thus, <math>F(12) = \frac{1}{2}(12)(11) = \boxed{066}</math>. | By the given, <math>F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12</math>. Thus, <math>F(12) = \frac{1}{2}(12)(11) = \boxed{066}</math>. | ||
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== See also == | == See also == |
Latest revision as of 22:10, 22 July 2008
Problem
Let be a polynomial such that
and
for
such that both sides are defined. Find
.
Solution
Combining denominators and simplifying,
It becomes obvious that
, for some constant
, matches the definition of the polynomial. To prove that
must have this form, note that
Since and
divides the right side of the equation,
and
divides the left side of the equation. Thus
divides
, so
divides
.
It is easy to see that is a quadratic, thus
as desired.
By the given, . Thus,
.
See also
Mock AIME 1 2007-2008 (Problems, Source) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |