2018 AIME II Problems/Problem 3

Problem

Find the sum of all positive integers $b < 1000$ such that the base-$b$ integer $36_{b}$ is a perfect square and the base-$b$ integer $27_{b}$ is a perfect cube.

Solution 1

The first step is to convert $36_{b}$ and $27_{b}$ into base-10 numbers. Then, we can write $36_{b}$ $= 3b + 6$ and $27_{b}$ $= 2b + 7$. It should also be noted that $8 \leq b < 1000$.

Because there are less perfect cubes than perfect squares for the restriction we are given on $b$, it is best to list out all the perfect cubes. Since the maximum $b$ can be is 1000 and $2$$1000 + 7 = 2007$, we can list all the perfect cubes less than 2007.

Now, $2b + 7$ must be one of $3^3, 4^3, ... , 12^3$. However, $2b + 7$ will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to $3^3, 5^3, 7^3, 9^3$, and $11^3$.

Because $3b + 6$ is a perfect square and is clearly divisible by 3, it must be divisible by 9, so $b + 2$ is divisible by 3. Thus the cube, which is $2b + 7 = 2(b + 2) + 3$, must also be divisible by 3. Therefore, the only cubes that $2b + 7$ could potentially be now are $3^3$ and $9^3$.

We need to test both of these cubes to make sure $3b + 6$ is a perfect square.

If we set $3^3 (27)$ equal to $2b + 7$, $b = 10$. If we plug this value of b into $3b + 6$, the expression equals $36$, which is indeed a perfect square.

If we set $9^3 (729)$ equal to $2b + 7$, $b = 361$. If we plug this value of b into $3b + 6$, the expression equals $1089$, which is $33^2$.

We have proven that both $b = 10$ and $b = 361$ are the only solutions, so $10 + 361 =$ $\boxed{371}$.

Solution 2

The conditions are: \[3b+6 = n^2\] \[2b+7 = m^3\] We can see $n$ is multiple is 3, so let $n=3k$, then $b= 3k^2-2$. Substitute $b$ into second condition and we get $m^3=3(2k^2+1)$. Now we know $m$ is both a multiple of 3 and odd. Also, $m$ must be smaller than 13 for $b$ to be smaller than 1000. So the only two possible values for $m$ are 3 and 9. Test and they both work. The final answer is $10 + 361 =$ $\boxed{371}$. -Mathdummy

Solution 3

As shown above, let \[3b+6 = n^2\] \[2b+7 = m^3\] such that \[6b+12=2n^2\] \[6b+21=3m^3\]

Subtracting the equations we have \[3m^3-2n^2=9 \implies 3m^3-2n^2-9=0.\]

We know that $m$ and $n$ both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solution $m$ must divide $9$ by the Rational Root Theorem.

We can instantly know $m \neq -9,-3,-1,1$ since those will have negative solutions.

When $m=3$ we have $n=6$, so then $b=10$

When $m=9$ we have $n=33$, so then $b=361$

Therefore, the sum of all possible values of $b$ is \[10+361=\boxed{371}.\]

See Also

2018 AIME II (ProblemsAnswer KeyResources)
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

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