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2) The least common mutiple of x and 9 is 45: x is 5, 15, or 45. nsf.

the same. Sufficient

Hey, guys, is it rule? LCM(a,b,c)=LCM(LCM(a,b),c) ?

If yes, we can use it: 1) LCM(x,6,9) = LCM(30,9)=90 2) LCM(x,6,9) = LCM(45,6)=90

D

Exactly Walker, this is the short cut rule. LCM(a,b,c)=LCM(LCM(a,b),c).

From stmt1, it LCM of 2 numbers(x,6) is given as 30. So LCM 3 numbers is LCM(30,9). Sufficient.

From stmt2, again same explanation as stmt1.

Initially I tried to find what is x from stmt 1 and then find the LCM of 3 numbers. THis is time consuming when the ans can be picked in less than 30 sec.

I raised this question, since I wanted to know in how ways can people solve the question.

As GMAT TIGER did, trying to find the x value rather than finding the LCM.

No offense, just analysing where all there are possibilities that things can go wrong and that we should be cautious.

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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.

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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

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