Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 4"
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===Original statement=== | ===Original statement=== | ||
− | Let <math> | + | Let <math>n</math> be the smallest positive integer for which there exist positive real numbers <math>a</math> and <math>b</math> such that <math>(a+bi)^n=(a-bi)^n</math>. Compute <math>\frac{b^2}{a^2}</math>. |
==Solution== | ==Solution== | ||
− | 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>. |
− | --- | + | ==See Also== |
+ | {{Mock AIME box|year=2006-2007|n=2|num-b=3|num-a=5}} | ||
− | + | [[Category:Intermediate Algebra Problems]] | |
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− | [[Category:Intermediate |
Latest revision as of 10:50, 4 April 2012
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
.
See Also
Mock AIME 2 2006-2007 (Problems, Source) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |