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

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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>.
 
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.
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Only the four-tuple <math>(2005,0,2005,0)</math> satisfies the equation, so the answer is <math>\boxed{1} \Rightarrow \mathrm{(B)}</math>.
  
 
== See also ==
 
== See also ==
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* [[2005 AMC 12B Problems]]
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{{AMC12 box|year=2005|ab=B|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 20:18, 21 December 2020

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, so the answer is $\boxed{1} \Rightarrow \mathrm{(B)}$.

See also

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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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