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

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==Solution 1==
 
==Solution 1==
  
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The first step is to convert <math>36_{b}</math> and <math>27_{b}</math> into base-10 numbers. Then, we can write <math>36_{b}</math> <math>= 3b + 6</math> and <math>27_{b}</math> <math>= 2b + 7</math>. It should also be noted that <math>8 \leq b < 1000</math>.
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The first step is to convert <math>36_{b}</math> and <math>27_{b}</math> into base-10 numbers. Then, we can write <cmath>36_{b} = 3b + 6</cmath> and <cmath>27_{b} = 2b + 7</cmath>. It should also be noted that <math>8 \leq b < 1000</math>.
  
 
Because there are less perfect cubes than perfect squares for the restriction we are given on <math>b</math>, it is best to list out all the perfect cubes. Since the maximum <math>b</math> can be is 1000 and <math>2</math> • <math>1000 + 7 = 2007</math>, we can list all the perfect cubes less than 2007.
 
Because there are less perfect cubes than perfect squares for the restriction we are given on <math>b</math>, it is best to list out all the perfect cubes. Since the maximum <math>b</math> can be is 1000 and <math>2</math> • <math>1000 + 7 = 2007</math>, we can list all the perfect cubes less than 2007.
  
−
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 <cmath>3^3, 4^3, ... , 12^3</cmath>. 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 <cmath>3^3, 5^3, 7^3, 9^3\text{, and }11^3</cmath>.
  
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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>.
+
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 <cmath>2b + 7 = 2(b + 2) + 3</cmath>, 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.
  
−
If we set <math>3^3 (27)</math> equal to <math>2b + 7</math>, <math>b = 10</math>. If we plug this value of b into <math>3b + 6</math>, the expression equals <math>36</math>, which is indeed a perfect square.
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<math>\textbf{Case 1:}</math> If we set <cmath>3^3 = 2b + 7</cmath>so <cmath>b = 10</cmath>. If we plug this value of b into <math>3b + 6</math>, the expression equals <math>36</math>, which is indeed a perfect square.
  
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If we set <math>9^3 (729)</math> equal to <math>2b + 7</math>, <math>b = 361</math>. If we plug this value of b into <math>3b + 6</math>, the expression equals <math>1089</math>, which is <math>33^2</math>.
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<math>\textbf{Case 2:}</math> If we set <cmath>9^3 = 2b + 7</cmath>so <cmath>b = 361</cmath>. If we plug this value of b into <math>3b + 6</math>, the expression equals <math>1089</math>, which is <math>33^2</math>.
 +
 
 +
We have proven that both <math>b = 10</math> and <math>b = 361</math> are the only solutions, so <cmath>10 + 361 = \boxed{371}</cmath>
  
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We have proven that both <math>b = 10</math> and <math>b = 361</math> are the only solutions, so <math>10 + 361 =</math> <math>\boxed{371}</math>.
 
 
==Solution 2==
 
==Solution 2==
 
The conditions are:
 
The conditions are:

Latest revision as of 01:28, 23 December 2023

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 \[3^3, 5^3, 7^3, 9^3\text{, and }11^3\].

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.

$\textbf{Case 1:}$ If we set \[3^3 = 2b + 7\]so \[b = 10\]. If we plug this value of b into $3b + 6$, the expression equals $36$, which is indeed a perfect square.

$\textbf{Case 2:}$ If we set \[9^3 = 2b + 7\]so \[b = 361\]. 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 (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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