Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"
(→=Solution 1) |
|||
Line 26: | Line 26: | ||
<cmath>6(6^2 - 3y) = 87 - 3x</cmath> | <cmath>6(6^2 - 3y) = 87 - 3x</cmath> | ||
<cmath>26 = x + y</cmath>. | <cmath>26 = x + y</cmath>. | ||
− | + | Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>. | |
Then <math>\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = \frac{69/7}{113/7} = \frac{69}{113}</math>. So, <math>m + n = 69 + 113 = \boxed{182}</math>. | Then <math>\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = \frac{69/7}{113/7} = \frac{69}{113}</math>. So, <math>m + n = 69 + 113 = \boxed{182}</math>. | ||
<baker77> | <baker77> |
Revision as of 11:19, 11 October 2019
Solution 1
For convenience, let's use instead of
. Define a polynomial
such that
. Let
and
. Then, our polynomial becomes
.
Note that we want to compute
.
From the given information, we know that the coefficient of the term is
, and we also know that
, or in other words,
. By Newton's Sums (since we are given
), we also find that
. Solving this system, we find that
. Thus,
, so our final answer is
.
Solution 2
Let ,
, and
. Then our system becomes
.
Since
, this equation becomes
.
.
Since
, this equation becomes
.
We will now use these equations to solve the problem. Let
, and
. Then we have
.
Our solutions are
and
.
Then . So,
.
<baker77>