Difference between revisions of "2005 AMC 12B Problems/Problem 17"

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== Solution ==
 
== Solution ==
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Using the laws of [[logarithms]], the given equation becomes
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<cmath>\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005</cmath>
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<cmath>\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005</cmath>
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<cmath>\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}</cmath>
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As <math>a,b,c,d</math> must all be rational, and there are no powers of <math>3</math> or <math>7</math> in <math>10^{2005}</math>, <math>b=d=0</math>. Then <math>2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005</math>.
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Only the four-tuple <math>(2005,0,2005,0)</math> satisfies the equation.
  
 
== See also ==
 
== See also ==
 
* [[2005 AMC 12B Problems]]
 
* [[2005 AMC 12B Problems]]

Revision as of 17:29, 21 February 2010

Problem

How many distinct four-tuples $(a,b,c,d)$ of rational numbers are there with

\[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]

$\mathrm{(A)}\ 0      \qquad \mathrm{(B)}\ 1      \qquad \mathrm{(C)}\ 17     \qquad \mathrm{(D)}\ 2004   \qquad \mathrm{(E)}\ \text{infinitely many}$

Solution

Using the laws of logarithms, the given equation becomes

\[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\] \[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\] \[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]

As $a,b,c,d$ must all be rational, and there are no powers of $3$ or $7$ in $10^{2005}$, $b=d=0$. Then $2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005$.

Only the four-tuple $(2005,0,2005,0)$ satisfies the equation.

See also