Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AIME II Problems/Problem 5"

(→‎Solution)
(→‎Solution II)
Line 13: Line 13:
  
 
==Solution II ==
 
==Solution II ==
−
Let <math>a=log_a b</math>. Then our equation becomes <math>a+\frac{1}{a}=5</math>. Multiplying through by <math>a</math> and solving the quadratic gives us <math>a=2</math> or <math>a=3</math>. Hence <math>a^2=b</math> or <math>a^3=b</math>.   
+
Let <math>a=log_a b</math>. Then our equation becomes <math>a+\frac{6}{a}=5</math>. Multiplying through by <math>a</math> and solving the quadratic gives us <math>a=2</math> or <math>a=3</math>. Hence <math>a^2=b</math> or <math>a^3=b</math>.   
  
 
For the first case <math>a^2=b</math>, <math>a</math> can range from 2 to 44, a total of 43 values.
 
For the first case <math>a^2=b</math>, <math>a</math> can range from 2 to 44, a total of 43 values.

Revision as of 15:51, 14 June 2011

Problem

Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$

Solution

The equation can be rewritten as $\frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5$ Multiplying through by $\log a \log b$ and factoring yields $(\log b - 3\log a)(\log b - 2\log a)=0$. Therefore, $\log b=3\log a$ or $\log b=2\log a$, so either $b=a^3$ or $b=a^2$.

  • For the case $b=a^2$, note that $44^2=1936$ and $45^2=2025$. Thus, all values of $a$ from $2$ to $44$ will work.
  • For the case $b=a^3$, note that $12^3=1728$ while $13^3=2197$. Therefore, for this case, all values of $a$ from $2$ to $12$ work.

There are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= \boxed{054}$.

Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs $(a,b)$, and not for the number of possible values of $b$. Were the problem to ask for the number of possible values of $b$, the values of $b^6$ under $2005$ would have to be subtracted, which would just be $2$ values: $2^6$ and $3^6$.


Solution II

Let $a=log_a b$. Then our equation becomes $a+\frac{6}{a}=5$. Multiplying through by $a$ and solving the quadratic gives us $a=2$ or $a=3$. Hence $a^2=b$ or $a^3=b$.

For the first case $a^2=b$, $a$ can range from 2 to 44, a total of 43 values. For the second case $a^3=b$, $a$ can range from 2 to 12, a total of 11 values.


Thus the total number of possible values is $43+11=\boxed{54}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_II_Problems/Problem_5&oldid=39543"