2005 AMC 12B Problems/Problem 17: Difference between revisions
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== Solution == | == Solution == | ||
Using the laws of [[logarithms]], the given equation becomes | |||
<cmath>\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005</cmath> | |||
<cmath>\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005</cmath> | |||
<cmath>\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}</cmath> | |||
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>. | |||
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 
of rational numbers are there with
![\[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]](http://latex.artofproblemsolving.com/9/1/5/9153c41adf40f93e619e0362b706aa5160874e9c.png)
Solution
Using the laws of logarithms, the given equation becomes
![\[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\]](http://latex.artofproblemsolving.com/9/b/3/9b367a566b912c5625212b4998ae7e4acb4a4918.png)
![\[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\]](http://latex.artofproblemsolving.com/3/4/b/34b1fdf053c6b026b0976f3f18c9a30db5ee001a.png)
![\[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]](http://latex.artofproblemsolving.com/8/5/9/859b1435302b7be43757fbe0695894a5fc73e38b.png)
As 
must all be rational, and there are no powers of 
or 
in 
, 
. Then 
.
Only the four-tuple 
satisfies the equation.
