Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 12B Problems/Problem 23"

($LaTeX$-ified)
m (→‎Solution)
Line 15: Line 15:
  
 
== Solution ==
 
== Solution ==
−
Call <math>x + y = s</math> and <math>x^2 + y^2 = t</math>.  Then, we note that log(s)=z which implies that <math>\log(10s) = z+1= \log(t)</math>.  Therefore, <math>t=10s</math>.  Let us note that <math>x^3 + y^3 = \frac[3st}{2}-\frac{s^2}{2} = s(15s-\frac{s^2}{2})</math>.  Since <math>s = 10^z</math>, we find that <math>x^3 + y^3 = 15\times10^2 - (1/2)\times10^3</math>. Thus, <math>a+b = \frac{29}{2}</math>.  <math>\boxed{\text{B}}</math> is the answer.
+
Call <math>x + y = s</math> and <math>x^2 + y^2 = t</math>.  Then, we note that <math>\log(s)=z</math> which implies that <math>\log(10s) = z+1= \log(t)</math>.  Therefore, <math>t=10s</math>.  Let us note that <math>x^3 + y^3 = \frac{3st}{2}-\frac{s^2}{2} = s(15s-\frac{s^2}{2})</math>.  Since <math>s = 10^z</math>, we find that <math>x^3 + y^3 = 15\times10^2 - (1/2)\times10^3</math>. Thus, <math>a+b = \frac{29}{2}</math>.  <math>\boxed{\text{B}}</math> is the answer.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2005|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2005|ab=B|num-b=22|num-a=24}}

Revision as of 13:08, 3 June 2011

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 $x^3 + y^3 = \frac{3st}{2}-\frac{s^2}{2} = s(15s-\frac{s^2}{2})$. 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.

See Also

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12B_Problems/Problem_23&oldid=39057"