2005 AIME II Problems/Problem 5
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.
and 
. Thus, all values of 
from 
to 
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 
and 
.
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 | ||
