Difference between revisions of "2008 AIME II Problems/Problem 15"
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where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n =\boxed{181}</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done. | where <math>k = 0,1,2,..</math>. So we can (with very little effort) obtain the following: <math>(k,a_k = 2n) = (0,2),(1,26),(2,362),(3,5042)</math>. It is an AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by induction. Checking <math>2n = 362\implies n =\boxed{181}</math> in the second condition works (we know <math>b_k</math> is odd so we don't need to find <math>m</math>). So we're done. | ||
− | === Solution 3 ( | + | === Solution 3 (BIG BAYUS) === |
Let us generate numbers <math>1</math> to <math>1000</math> for the second condition, for squares. We know for <math>N</math> to be integer, the squares must be odd. So we generate <math>N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it is AIME problem. Now take the first criterion, let <math>a</math> be the smaller consecutive cube. We then get: | Let us generate numbers <math>1</math> to <math>1000</math> for the second condition, for squares. We know for <math>N</math> to be integer, the squares must be odd. So we generate <math>N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it is AIME problem. Now take the first criterion, let <math>a</math> be the smaller consecutive cube. We then get: |
Revision as of 01:19, 2 January 2019
Problem
Find the largest integer satisfying the following conditions:
- (i)
can be expressed as the difference of two consecutive cubes;
- (ii)
is a perfect square.
Contents
[hide]Solution
Solution 1
Write , or equivalently,
.
Since and
are both odd and their difference is
, they are relatively prime. But since their product is three times a square, one of them must be a square and the other three times a square. We cannot have
be three times a square, for then
would be a square congruent to
modulo
, which is impossible.
Thus is a square, say
. But
is also a square, say
. Then
. Since
and
have the same parity and their product is even, they are both even. To maximize
, it suffices to maximize
and check that this yields an integral value for
. This occurs when
and
, that is, when
and
. This yields
and
, so the answer is
.
Solution 2
Suppose that the consecutive cubes are and
. We can use completing the square and the first condition to get:
where
and
are non-negative integers. Now this is a Pell equation, with solutions in the form
. However,
is even and
is odd. It is easy to see that the parity of
and
switch each time (by induction). Hence all solutions to the first condition are in the form:
where
. So we can (with very little effort) obtain the following:
. It is an AIME problem so it is implicit that
, so
. It is easy to see that
is strictly increasing by induction. Checking
in the second condition works (we know
is odd so we don't need to find
). So we're done.
Solution 3 (BIG BAYUS)
Let us generate numbers to
for the second condition, for squares. We know for
to be integer, the squares must be odd. So we generate
.
cannot exceed
since it is AIME problem. Now take the first criterion, let
be the smaller consecutive cube. We then get:
Now we know either or
must be factor of
, hence
or
. Only
satisfy this criterion. Testing each of the numbers in the condition yields
as the largest that fits both, thus answer
.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last problem | |
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.