2005 AMC 12B Problems/Problem 23

Problem

Let $S$ be the set of ordered triples $(x,y,z)$ of real numbers for which

$\log_{10}(x+y) = z \text{ and } \log_{10}(x^{2}+y^{2}) = z+1.$ 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 $a+b?$

$\textbf{(A)}\ \frac {15}{2} \qquad \textbf{(B)}\ \frac {29}{2} \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ \frac {39}{2} \qquad \textbf{(E)}\ 24$

Solution 1

Let $x + y = s$ and $x^2 + y^2 = t$ . Then, $\log(s)=z$ implies $\log(10s) = z+1= \log(t)$ ,so $t=10s$ . Therefore, $x^3 + y^3 = s*\frac{3t-s^2}{2} = s(15s-\frac{s^2}{2})$ . Since $s = 10^z$ , we find that $x^3 + y^3 = 15\times10^{2z} - (1/2)\times10^{3z}$ . Thus, $a+b = \frac{29}{2}$ , $\boxed{B}$

Solution 2

First, remember that $x^3 + y^3$ factors to $(x + y) (x^2 - xy + y^2)$ . By the givens, $x + y = 10^z$ and $x^2 + y^2 = 10^{z + 1}$ . These can be used to find $xy$ : $(x + y)^2 = 10^{2z}$ $x^2 + 2xy + y^2 = 10^{2z}$ $2xy = 10^{2z} - 10^{z + 1}$ $xy = \frac{10^{2z} - 10^{z + 1}}{2}$

Therefore, $x^3 + y^3 = a \cdot 10^{3z} + b \cdot 10^{2z} = 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)$ $= 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)$ $= 10^{2z + 1} - \frac{10^{3z} - 10^{2z + 1}}{2}$ $= -\frac{1}{2} \cdot 10^{3z} + \frac{3}{2} \cdot 10^{2z + 1}$ $= -\frac{1}{2} \cdot 10^{3z} + 15 \cdot 10^{2z}.$

It follows that $a = -\frac{1}{2}$ and $b = 15$ , thus $a + b = \frac{29}{2}.$

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2005 AMC 12B Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See Also