Difference between revisions of "2005 AIME II Problems/Problem 5"
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Determine the number of [[ordered pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math> | Determine the number of [[ordered pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math> | ||
| − | == Solution | + | == Solution 1 == |
The equation can be rewritten as <math> \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \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>, so either <math> b=a^3 </math> or <math> b=a^2 </math>. | The equation can be rewritten as <math> \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \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>, so 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 <math>a</math> from <math>2</math> to <math>44</math> will work. | *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 <math>a</math> from <math>2</math> to <math>44</math> will work. | ||
Revision as of 13:27, 27 November 2019
Contents
Problem
Determine the number of ordered pairs 
of integers such that 
and 
Solution 1
The equation can be rewritten as 
Multiplying through by 
and factoring yields 
. Therefore, 
or 
, so either 
or 
.
- For the case
, note that 
and 
. Thus, all values of 
from 
to 
will work. - For the case
, note that 
while 
. Therefore, for this case, all values of from to 
work.
and 
. Thus, all values of 
will work.
while 
. Therefore, for this case, all values of 
work.There are 
possibilities for the square case and 
possibilities for the cube case. Thus, the answer is 
.
Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs , and not for the number of possible values of 
. Were the problem to ask for the number of possible values of , the values of 
under 
would have to be subtracted, which would just be 
values: 
and 
. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)
Solution 2
Let 
. Then our equation becomes 
. Multiplying through by 
and solving the quadratic gives us 
or 
. Hence 
or 
.
For the first case , 
can range from 2 to 44, a total of 43 values.
For the second case , can range from 2 to 12, a total of 11 values.
Thus the total number of possible values is 
.
See also
| 2005 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
