Difference between revisions of "2005 AIME II Problems/Problem 5"
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| + | Note that the equation can be rewritten as: <math> \frac{\log b}{\log a}+6\frac{\log a}{\log b}=5 </math> or <math> \frac{(\log b)^2+6(\log a)^2}{\log a * \log b}=5 </math> Multiplying through by <math> \log a * \log b </math> and factoring yields <math> (\log b - 3\log a)(\log b - 2\log a)=0 </math>. Therefore, <math> \log b=3\log a </math> or <math> \log b=2\log a </math>. Therefore, either <math> b=a^3 </math> or <math> b=a^2 </math>. For the case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of a from 2 to 44 will work. For the case <math> b=a^3 </math>, note that <math> 12^3=1728 </math> while <math> 13^3=2197 </math>. Therefore, for this case, all values of a from 2 to 12 work. There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11=054 </math>. | ||
== See Also == | == See Also == | ||
Revision as of 22:13, 12 March 2007
Problem
Determine the number of ordered pairs 
of integers such that 
and 
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Note that the equation can be rewritten as: 
or 
Multiplying through by 
and factoring yields 
. Therefore, 
or 
. Therefore, either 
or 
. For the case , note that 
and 
. Thus, all values of a from 2 to 44 will work. For the case , note that 
while 
. Therefore, for this case, all values of a from 2 to 12 work. There are 
possibilities for the square case and 
possibilities for the cube case. Thus, the answer is 
.
