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

Call $x + y = s$ and $x^2 + y^2 = t$ . Then, we note that $\log(s)=z$ which implies that $\log(10s) = z+1= \log(t)$ . Therefore, $t=10s$ . Let us note that $x^3 + y^3 = \frac{3st}{2}-\frac{s^3}{2} = s(15s-\frac{s^2}{2})$ . Since $s = 10^z$ , we find that $x^3 + y^3 = 15\times10^2 - (1/2)\times10^3$ . Thus, $a+b = \frac{29}{2}$ . $\boxed{\text{B}}$ is the answer.

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2005 AMC 12B Problems/Problem 23

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