Difference between revisions of "2018 AIME II Problems/Problem 3"
Jerry122805 (talk | contribs) (→Solution 2) |
Jerry122805 (talk | contribs) (→Solutino 3) |
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As shown above, let | As shown above, let | ||
<cmath>3b+6 = n^2</cmath> | <cmath>3b+6 = n^2</cmath> | ||
− | <cmath>2b+7 = m^3</cmath> such that <cmath>6b+12=2n^2</cmath> | + | <cmath>2b+7 = m^3</cmath> such that <cmath>6b+12=2n^2</cmath> <cmath>6b+21=3m^3</cmath> |
==See Also== | ==See Also== |
Revision as of 19:33, 6 March 2020
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 1
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
.
Solution 2
The conditions are:
We can see
is multiple is 3, so let
, then
. Substitute
into second condition and we get
. Now we know
is both a multiple of 3 and odd. Also,
must be smaller than 13 for
to be smaller than 1000. So the only two possible values for
are 3 and 9. Test and they both work. The final answer is
.
-Mathdummy
Solutino 3
As shown above, let
such that
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.