Difference between revisions of "2005 AIME II Problems/Problem 5"
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Thus the total number of possible values is <math>43+11=\boxed{54}</math>. | Thus the total number of possible values is <math>43+11=\boxed{54}</math>. | ||
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+ | ==Solution 3(similar to solution 2)== | ||
+ | Using the change of base formula on the second equation to change to base <math>a</math>, we get <math>\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}</math>. If we substitute <math>x</math> for <math>\log_a(b)</math>, we get <math>x + \frac{6}{x}</math>. Multiplying by <math>x</math> on both sides and solving, we get <math>x=3,2</math>. Substituting back in, we get <math>\log_a(b) = 3,2</math>. That means <math>a^3 = b</math> or <math>a^2 = b</math>. Since <math>b \leq 2005</math>, we can see that for the cubed case, the maximum <math>a</math> can be without exceeding 2005 is 12(because <math>13^3 = 2197</math>) and for the squared case it can be a maximum of 44. Since <math>a \neq 1</math>, the number of values is <math>(44-1)+(12-1) = \boxed{54}</math>. | ||
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+ | ~idk12345678 | ||
== See also == | == See also == |
Latest revision as of 19:15, 4 April 2024
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.
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 .
Solution 3(similar to solution 2)
Using the change of base formula on the second equation to change to base , we get
. If we substitute
for
, we get
. Multiplying by
on both sides and solving, we get
. Substituting back in, we get
. That means
or
. Since
, we can see that for the cubed case, the maximum
can be without exceeding 2005 is 12(because
) and for the squared case it can be a maximum of 44. Since
, the number of values is
.
~idk12345678
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.