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by Rich@VeritasPrep » Thu Jul 01, 2010 8:16 am
1. Is n divisible by 108?

(1) n is divisible by 12
(2) n is divisible by 9

This one is fairly straightforward if you keep in mind that we can show insufficiency by finding both yes and no answers to the prompt.

36 is a number that's divisible by both 12 and 9, but it is not divisible by 108.

108 is divisible by both 12 and 9, and it is divisible by 108.

Done and done. Answer: E
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by er.twi.fb » Thu Jul 01, 2010 8:23 am
I was counting on [C] because if you combine both of them, then it tells n is divisible by 108.

1>n is divisible by 2^2*3
2>n is divisible by 3^2

1+2>n is divisible by 2^2*3^3(=108)

Your answer is correct, but I am still confused.
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by Rich@VeritasPrep » Thu Jul 01, 2010 8:28 am
Be careful ... just because a value is divisible by two numbers does NOT necessarily mean that it is divisible by the product of those numbers.

For example, 4 is divisible by both 2 and 4. Does that mean 4 is divisible by 8? Nope.

In this case, the problem is that 12 and 9 share a common factor, namely 3, and thus their Least Common Multiple will NOT be 12*9 = 108. The LCM of 12 and 9 is instead 36.

You can use prime factorization (as you started to do) to show this:

9 = 3^2
12 = 2^2 * 3

The LCM will have 2^2 (since 12 carries the highest power of 2) and 3^2 (since 9 carries the highest power of 3). You don't multiply the 3^2 from 9 and the 3 from 12 to get 3^3, because two powers of 3 is all you need as part of an LCM for both numbers.

That means the LCM is 2^2 * 3^2 = 36.

Make sense?
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by er.twi.fb » Thu Jul 01, 2010 8:40 am
Thanks, it totally makes sense to me now.
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