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>. | ||
− | 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 21:44, 24 March 2018
Problem
Find the sum of all positive integers such that the base-
integer
is a perfect square and the base-
integer
is a perfect cube.
Solution
The first step is to convert and
into base-10 numbers. Then, we can write
and
. It should also be noted that
.
Because there are less perfect cubes than perfect squares for the restriction we are given on , it is best to list out all the perfect cubes. Since the maximum
can be is 1000 and
•
, we can list all the perfect cubes less than 2007.
Now, must be one of
. However,
will always be odd, so we can eliminate the cubes of the even numbers and change our list of potential cubes to
, and
.
Because is a perfect square and is clearly divisible by 3, it must be divisible by 9, so
is divisible by 3. Thus the cube, which is
, must also be divisible by 3. Therefore, the only cubes that
could potentially be now are
and
.
We need to test both of these cubes to make sure is a perfect square.
If we set equal to
,
. If we plug this value of b into
, the expression equals
, which is indeed a perfect square.
If we set equal to
,
. If we plug this value of b into
, the expression equals
, which is
.
We have proven that both and
are the only solutions, so
.
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.