by RonPurewal Fri Jan 09, 2009 6:48 am
you absolutely must know the meaning of "least common multilple", as it's a very important concept.
the least common multiple of x and y should be thought of in two separate ways:
(1) it's the smallest number that has both x and y as factors, or, equivalently, into which both x and y divide evenly.
this is the simplistic, grade-school approach. it works fine when you have small, concrete numbers to deal with - for instance, the least common multiple of 6 and 4 is 12 - but it falters with larger numbers, and falls flat on its face with variables.
(2) it's the number made from all the primes appearing in the prime factorization of either x or y, raised to the higher of the powers to which it's raised in those prime factorizations.
this is extremely awkward to put in words, so an example will suffice:
the LCM of (2^6)(3^5)(7^4) and (2^4)(3^7)(5)
is
(2^6)(3^7)(5)(7^4).
if you want the lcm of more than 2 numbers, generalize these accordingly.
--
use definition (2) here, since you have unknowns to deal with.
statement (1):
the least common multiple of x and (2)(3) is (2)(3)(5).
this means:
we KNOW x contains exactly one 5, because otherwise that 5 wouldn't be in the lcm.
x MAY contain a 2, a 3, both, or neither; any of these possibilities would yield the (2)(3) in the lcm. note that x cannot contain more than one 2 or 3, as those powers would then go into the lcm.
x cannot contain any other primes, because those primes would have to appear in the lcm, and they don't.
given these facts, you know that the lcm of x, (3)(2), and (3^2) is (2)(3^2)(5). if you don't see why right away, run through the possibilities.
--
statement (2):
use the same kind of analysis as that used for statement (1).
the lcm of x and (3^2) is (3^2)(5).
therefore,
we KNOW that x contains exactly one 5.
x MAY contain no, one, or two 3's.
x CANNOT contain any other primes.
same sort of reasoning used above --> the lcm must be (2)(3^2)(5); sufficient
--
fun fact:
if you take the lcm of x and y, and then take the lcm of that number and z, you'll just get the lcm of x, y, and z. it doesn't matter which ones you look at pairwise first, since the biggest powers win anyway.
if you know this, then this problem's solution is immediate, because (1) means you want the lcm of 30 and 9, and (2) means you want the lcm of 45 and 6. both of those are single numerical values, so they must be sufficient.
