Problem:
Let [r,s] denote the least common multiple of positive integers r and s. Find the number of ordered triples (a,b,c) of positive integers for which [a,b]=1000,[b,c]=2000, and [c,a]=2000.
Solution:
Since both 1000 and 2000 are of the form 2m5n, the numbers a,b and c must also be of this form. More specifically,
a=2m15n1,b=2m25n2,c=2m35n3,(1)
where the mi and ni are non-negative integers for i=1,2,3.
Then, in view of the definition of [r,s], and since
[a,b]=2353,[b,c]=2453,[c,a]=2453,(2)
the following equalities must hold:
max{m1,m2}=3,max{m2,m3}=4,max{m3,m1}=4(3)
and
max{n1,n2}=3,max{n2,n3}=3,max{n3,n1}=3(4)
To satisfy (3), we must have m3=4, and either m1 or m2 must be 3, while the other one can take the values of 0,1,2 or 3. There are 7 such ordered triples, namely (0,3,4),(1,3,4),(2,3,4),(3,0,4), (3,1,4),(3,2,4) and (3,3,4).
To satisfy (4), two of n1,n2 and n3 must be 3, while the third one ranges through the values of 0,1,2 and 3. The number of such ordered triples is 10; they are (3,3,0),(3,3,1),(3,3,2),(3,0,3),(3,1,3),(3,2,3),(0,3,3),(1,3,3),(2,3,3) and (3,3,3).
Since the choice of (m1,m2,m3) is independent of the choice of (n1,n2,n3), they can be chosen in 7⋅10=70 different ways. This is the number of ordered triples (a,b,c) satisfying the given conditions.
The problems on this page are the property of the MAA's American Mathematics Competitions