Difference between revisions of "2018 AIME II Problems/Problem 3"

Revision as of 18:35, 6 March 2020

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

Solutino 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 solutino $m$ must divide $9$ by the Rational Root Theorem.

@@ Line 33: / Line 33: @@
 <cmath>2b+7 = m^3</cmath> such that <cmath>6b+12=2n^2</cmath> <cmath>6b+21=3m^3</cmath>
-Subtracting the equations we have <cmath>3m^3-2n^2=9 \implies 3m^2-2n^2-9=0</cmath>
+Subtracting the equations we have <cmath>3m^3-2n^2=9 \implies 3m^3-2n^2-9=0.</cmath>
+We know that <math>m</math> and <math>n</math> both have to be integers, because then the base wouldn't be an integer. Furthermore, any integer solutino <math>m</math> must divide <math>9</math> by the Rational Root Theorem.
 ==See Also==

2018 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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