2018 AIME II Problems/Problem 3
Problem
Find the sum of all positive integers such that the base- integer is a perfect square and the base- integer is a perfect cube.
Solution 1
The first step is to convert and into base-10 numbers. Then, we can write and . It should also be noted that .
Because there are less perfect cubes than perfect squares for the restriction we are given on , it is best to list out all the perfect cubes. Since the maximum can be is 1000 and • , we can list all the perfect cubes less than 2007.
Now, must be one of . However, will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to , and .
Because is a perfect square and is clearly divisible by 3, it must be divisible by 9, so is divisible by 3. Thus the cube, which is , must also be divisible by 3. Therefore, the only cubes that could potentially be now are and .
We need to test both of these cubes to make sure is a perfect square.
If we set equal to , . If we plug this value of b into , the expression equals , which is indeed a perfect square.
If we set equal to , . If we plug this value of b into , the expression equals , which is .
We have proven that both and are the only solutions, so .
Solution 2
The conditions are: We can see is multiple is 3, so let , then . Substitute into second condition and we get . Now we know is both a multiple of 3 and odd. Also, must be smaller than 13 for to be smaller than 1000. So the only two possible values for are 3 and 9. Test and they both work. The final answer is . -Mathdummy
Solution 3
As shown above, let such that
Subtracting the equations we have
We know that and both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution must divide by the Rational Root Theorem.
We can instantly know since those will have negative solutions.
When we have , so then
When we have , so then
Therefore, the sum of all possible values of is
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.