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

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== Solution ==
 
== Solution ==
 
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^3}{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^3}{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.
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 +
==Alternate Solution==
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First, remember that <math>x^3 + y^3</math> factors to <math>(x + y) (x^2 - xy + y^2)</math>. By the givens, <math>x + y = 10^z</math> and <math>x^2 + y^2 = 10^{z + 1}</math>. These can be used to find <math>xy</math>:
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<cmath>(x + y)^2 = 10^{2z}</cmath>
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<cmath>x^2 + 2xy + y^2 = 10^{2z}</cmath>
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<cmath>2xy = 10^{2z} - 10^{z + 1}</cmath>
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<cmath>xy = \frac{10^{2z} - 10^{z + 1}}{2}</cmath>
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 +
Therefore,
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<cmath>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)</cmath>
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<cmath>= 10^z\left(10^{z + 1} - \frac{10^{2z} - 10^{z + 1}}{2}\right)</cmath>
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<cmath>= 10^{2z + 1} - \frac{10^{3z} - 10^{2z + 1}}{2}</cmath>
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<cmath>= -\frac{1}{2} \cdot 10^{3z} + \frac{3}{2} \cdot 10^{2z}</cmath>
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<cmath>= -\frac{1}{2} \cdot 10^{3z} + 15 \cdot 10^{2z}.</cmath>
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 +
It follows that <math>a = -\frac{1}{2}</math> and <math>b = 15</math>, thus <math>a + b = \frac{29}{2}.</math>
  
 
== 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 20:41, 29 June 2013

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.

Alternate Solution

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

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

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
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