Difference between revisions of "2005 AMC 12B Problems/Problem 17"
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<cmath>\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005</cmath> | <cmath>\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005</cmath> | ||
| − | <cmath>\ | + | <cmath>\rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005</cmath> |
| − | <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>. | 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>. | ||
Revision as of 20:17, 21 December 2020
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/3/6/a/36a4d0d1fd9fa1a448a433a28fe52630bbf76155.png)
![\[\rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]](http://latex.artofproblemsolving.com/7/5/6/7565d05ca1a55069803e6c9fd1ac7f50ab6e8cf3.png)
As 
must all be rational, and there are no powers of 
or 
in 
, 
. Then 
.
Only the four-tuple 
satisfies the equation, so the answer is 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
