Difference between revisions of "1987 AIME Problems/Problem 7"
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== Problem == | == Problem == | ||
− | Let <math>\displaystyle [r,s]</math> denote the [[least common multiple]] of [[positive | + | Let <math>\displaystyle [r,s]</math> denote the [[least common multiple]] of [[positive integer]]s <math>\displaystyle r</math> and <math>\displaystyle s</math>. Find the number of [[ordered tuple | ordered triples]] <math>\displaystyle (a,b,c)</math> of positive integers for which <math>\displaystyle [a,b] = 1000</math>, <math>\displaystyle [b,c] = 2000</math>, and <math>\displaystyle [c,a] = 2000</math>. |
== Solution == | == Solution == | ||
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== See also == | == See also == | ||
{{AIME box|year=1987|num-b=6|num-a=8}} | {{AIME box|year=1987|num-b=6|num-a=8}} | ||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 21:01, 15 February 2007
Problem
Let denote the least common multiple of positive integers
and
. Find the number of ordered triples
of positive integers for which
,
, and
.
Solution
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 12 possible triples.
Since the exponents of 2 and 5 must satisfy these conditions independently, we have a total of possible valid triples.
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
.
c | a | b | solutions |
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31 |
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18 | |
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24 |
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 |