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Difference between revisions of "2018 AIME II Problems/Problem 3"
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Now, <math>2b + 7</math> must be one of <math>3^3, 4^3, ... , 12^3</math>. However, <math>2b + 7</math> will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to <math>3^3, 5^3, 7^3, 9^3</math>, and <math>11^3</math>.
 
Now, <math>2b + 7</math> must be one of <math>3^3, 4^3, ... , 12^3</math>. However, <math>2b + 7</math> will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to <math>3^3, 5^3, 7^3, 9^3</math>, and <math>11^3</math>.
  
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Because <math>3b + 6</math> is a perfect square and is clearly divisible by 3, the cube, which is <math>2b + 7</math>, must also be divisible by 3. Therefore, the only cubes that <math>2b + 7</math> could potentially be now are <math>3^3</math> and <math>9^3</math>.
+
Because <math>3b + 6</math> is a perfect square and is clearly divisible by 3, it must be divisible by 9, so <math>b + 2</math> is divisible by 3. Thus the cube, which is <math>2b + 7 = 2(b + 2) + 3</math>, must also be divisible by 3. Therefore, the only cubes that <math>2b + 7</math> could potentially be now are <math>3^3</math> and <math>9^3</math>.
  
 
We need to test both of these cubes to make sure <math>3b + 6</math> is a perfect square.
 
We need to test both of these cubes to make sure <math>3b + 6</math> is a perfect square.

Revision as of 20:44, 24 March 2018

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

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 $10 + 361 =$ $\boxed{371}$.

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

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