Difference between revisions of "2005 AIME II Problems/Problem 5"
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==Solution II == | ==Solution II == | ||
− | Let <math> | + | Let <math>k=\log_a b</math>. Then our equation becomes <math>k+\frac{6}{k}=5</math>. Multiplying through by <math>k</math> and solving the quadratic gives us <math>k=2</math> or <math>k=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 05:04, 14 August 2011
Contents
Problem
Determine the number of ordered pairs of integers such that and
Solution
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.
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 .
Solution II
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 |