Difference between revisions of "2005 AIME II Problems/Problem 5"
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Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>. Were the problem to ask for the number of possible values of <math>b</math>, the values of <math>b^6</math> under <math>2005</math> would have to be subtracted, which would just be <math>2</math> values: <math>2^6</math> and <math>3^6</math>. 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).) | Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>. Were the problem to ask for the number of possible values of <math>b</math>, the values of <math>b^6</math> under <math>2005</math> would have to be subtracted, which would just be <math>2</math> values: <math>2^6</math> and <math>3^6</math>. 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 | + | ==Solution 2 == |
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>. | 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>. | ||
Revision as of 13:27, 27 November 2019
Contents
Problem
Determine the number of ordered pairs 
of integers such that 
and 
Solution I
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. 
