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 | ||
+ | |||
+ | <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
Solution
Using the laws of logarithms, the given equation becomes
As must all be rational, and there are no powers of or in , . Then .
Only the four-tuple satisfies the equation.