Difference between revisions of "1987 AIME Problems/Problem 7"
Lazymathfans (talk | contribs) m (→Solution 1) |
|||
Line 8: | Line 8: | ||
Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 3</math>. Thus, at least two of <math>k, n, q</math> must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples: <math>(3, 3, 3)</math> and three possibilities of each of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>. | Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 3</math>. Thus, at least two of <math>k, n, q</math> must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples: <math>(3, 3, 3)</math> and three possibilities of each of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>. | ||
− | Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of <math>7 \cdot 10 = | + | Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of <math>7 \cdot 10 = 70</math> possible valid triples. |
==Solution 2== | ==Solution 2== |
Revision as of 22:53, 4 April 2016
Problem
Let denote the least common multiple of positive integers
and
. Find the number of ordered triples
of positive integers for which
,
, and
.
Contents
Solution 1
It's clear that we must have ,
and
for some nonnegative integers
. Dealing first with the powers of 2: from the given conditions,
,
. Thus we must have
and at least one of
equal to 3. This gives 7 possible triples
:
and
.
Now, for the powers of 5: we have . Thus, at least two of
must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples:
and three possibilities of each of the forms
,
and
.
Since the exponents of 2 and 5 must satisfy these conditions independently, we have a total of possible valid triples.
Solution 2
and
. By looking at the prime factorization of
,
must have a factor of
. If
has a factor of
, then there are two cases: either (1)
or
, or (2) one of
and
has a factor of
and the other a factor of
. For case 1, the other number will be in the form of
, so there are
possible such numbers; since this can be either
or
there are a total of
possibilities. For case 2,
and
are in the form of
and
, with
and
(if they were equal to 3, it would overlap with case 1). Thus, there are
cases.
If does not have a factor of
, then at least one of
and
must be
, and both must have a factor of
. Then, there are
solutions possible just considering
, and a total of
possibilities. Multiplying by three, as
, there are
. Together, that makes
solutions for
.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.