Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 4"
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− | Two [[complex number]]s are equal if and only if their [[real part]]s and [[imaginary part]]s are equal. Thus if <math>(a + bi)^1 = (a - bi)^1</math> we have <math>b = -b</math> so <math>b = 0</math>, not a positive number. If <math>(a + bi)^2 = (a - bi)^2</math> we have <math>2ab = - 2ab</math> so <math>4ab = 0</math> so <math>a = 0</math> or <math>b = 0</math>, again violating the givens. <math>(a + bi)^3 = (a -bi)^3</math> is equivalent to <math>a^3 - 3ab^2 = a^3 - 3ab^2</math> and <math>3a^2b - b^3 = -3a^2b + b^3</math>, which are true if and only if <math>3a^2b = b^3</math> so either <math>b = 0</math> or <math>3a^2 = b^2</math>. Thus <math>n = | + | Two [[complex number]]s are equal if and only if their [[real part]]s and [[imaginary part]]s are equal. Thus if <math>(a + bi)^1 = (a - bi)^1</math> we have <math>b = -b</math> so <math>b = 0</math>, not a positive number. If <math>(a + bi)^2 = (a - bi)^2</math> we have <math>2ab = - 2ab</math> so <math>4ab = 0</math> so <math>a = 0</math> or <math>b = 0</math>, again violating the givens. <math>(a + bi)^3 = (a -bi)^3</math> is equivalent to <math>a^3 - 3ab^2 = a^3 - 3ab^2</math> and <math>3a^2b - b^3 = -3a^2b + b^3</math>, which are true if and only if <math>3a^2b = b^3</math> so either <math>b = 0</math> or <math>3a^2 = b^2</math>. Thus <math>n = b^2/a^2 = 3</math>. |
Revision as of 11:29, 16 September 2006
Problem
Revised statement
Let and
be positive real numbers and
a positive integer such that
, where
is as small as possible and
. Compute
.
Original statement
Let be the smallest positive integer for which there exist positive real numbers
and
such that
. Compute
.
Solution
Two complex numbers are equal if and only if their real parts and imaginary parts are equal. Thus if we have
so
, not a positive number. If
we have
so
so
or
, again violating the givens.
is equivalent to
and
, which are true if and only if
so either
or
. Thus
.