Question

Let $S$ be the set of ordered triples $(x, y, z)$ of real numbers for which $\log_{10}(x + y) = z$ and $\log_{10}(x^2 + y^2) = z + 1.$ There are real numbers $a$ and $b$ such that for all ordered triples $(x, y, z)$ in $S$ we have $x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z}.$ What is the value of $a + b?$

$\dfrac{15}{2}$

$\dfrac{29}{2}$

$15$

$\dfrac{39}{2}$

$24$

Answer: B

Concepts:sum and difference of cubes logarithm symmetry (algebra)

Difficulty rating: 2110

Solution:

The conditions give $x + y = 10^z$ and $x^2 + y^2 = 10 \cdot 10^z.$ Then $\begin{aligned} &2xy = (x+y)^2 - (x^2+y^2) \\ &= 10^{2z} - 10 \cdot 10^z, \end{aligned}$ so $xy = \dfrac12\left(10^{2z} - 10 \cdot 10^z\right).$

Using $x^3 + y^3$ $= (x+y)^3 - 3xy(x+y),$ $\begin{aligned} &x^3 + y^3 = 10^{3z} \\ &\quad {}- \dfrac32\left(10^{2z} - 10 \cdot 10^z\right)10^z \\ &= -\dfrac12 \cdot 10^{3z} + 15 \cdot 10^{2z}. \end{aligned}$

So $a = -\dfrac12$ and $b = 15,$ giving $a + b = \dfrac{29}{2}.$ These coefficients are determined: setting $z=0$ gives real solutions $x,y=\dfrac{1\pm\sqrt{19}}{2},$ and their equation forces this same value of $a+b.$

Thus, the correct answer is B.

2005 AMC 12B Problem 23

Answer: B

Difficulty rating: 2110

Solution:

Problem 23 in Other Years