2005 AIME II Problems/Problem 5
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.
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 |
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