Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 2)
Line 24: Line 24:
 
<cmath>3b+6 = n^2</cmath>
 
<cmath>3b+6 = n^2</cmath>
 
<cmath>2b+7 = m^3</cmath>
 
<cmath>2b+7 = m^3</cmath>
−
We can see <math>n</math> is multiple is 3, so let <math>n=3k</math>, then <math>b= 3k^2-2</math>, substitute <math>b</math> into second condition we have <math>m^3=3(k^2+1)</math>. Now we know <math>m</math> is multiple of 3, and is odd. Also, <math>m</math> must be smaller than 15 for <math>b</math> to be smaller than 1000. So the only two possible values for <math>m</math> is 3 and 9. Test and they both works. The final answer is <math>10 + 361 =</math> <math>\boxed{371}</math>. -Mathdummy
+
We can see <math>n</math> is multiple is 3, so let <math>n=3k</math>, then <math>b= 3k^2-2</math>, substitute <math>b</math> into second condition we have <math>m^3=3(2k^2+1)</math>. Now we know <math>m</math> is multiple of 3, and is odd. Also, <math>m</math> must be smaller than 15 for <math>b</math> to be smaller than 1000. So the only two possible values for <math>m</math> is 3 and 9. Test and they both works. The final answer is <math>10 + 361 =</math> <math>\boxed{371}</math>. -Mathdummy
  
 
==See Also==
 
==See Also==

Revision as of 18:51, 11 March 2019

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 we have $m^3=3(2k^2+1)$. Now we know $m$ is multiple of 3, and is odd. Also, $m$ must be smaller than 15 for $b$ to be smaller than 1000. So the only two possible values for $m$ is 3 and 9. Test and they both works. The final answer is $10 + 361 =$ $\boxed{371}$. -Mathdummy

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_3&oldid=104282"