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by Stuart@KaplanGMAT » Tue Jan 29, 2008 10:52 am
If x is a positive integer, what's the LCM of x, 6 and 9?

(1) The LCM of x and 6 is 30.

We can attack this statement by brute force or by using our knowledge of principles.

If the LCM of x and 6 is 30, then x could be 5, 10, 15 or 30. If you try out all 4 of those possibilities, you'll get a LCM of x, 6 and 9 of 90. Since we get the same answer every time, (1) is sufficient.

Alternatively, if the LCM Of x and 6 is 30, that means that x must have a prime factor of 5 (since the "5" in 30 has to come from either x or 6, and it doesn't come from 6). So, we need a "5" from x, a "2" and "3" from 6 and another "3" from 9. 2*3*3*5 = 90. Sufficient.

(2) the LCM of x and 9 is 45.

Same solution as above!

The possibly values for x are 5, 15 and 45. In all 3 cases, the LCM of x, 6 and 9 is 90. Sufficient.

Alternatively, if the LCM of x and 9 is 45, once again the prime factor of "5" has to come from the x. We're going to end up with 2*3*3*5 yet again. Sufficient.

Both statements are sufficient alone: choose (D).
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by nadontheway » Tue Jan 29, 2008 12:41 pm
Thanks Stuart. I see your point but could you elaborate on the concept of LCM?

x= 5
6= 3*2
9=3*3

to get the LCM of x, 6 and 9, we pick 5 from x, 2 from 6 and 3*3 from 9 that gives us 90. Is it the way to proceed?
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by Stuart@KaplanGMAT » Tue Jan 29, 2008 2:07 pm
nadontheway wrote:Thanks Stuart. I see your point but could you elaborate on the concept of LCM?

x= 5
6= 3*2
9=3*3

to get the LCM of x, 6 and 9, we pick 5 from x, 2 from 6 and 3*3 from 9 that gives us 90. Is it the way to proceed?
LCM stands for lowest common multiple.

The LCM of two (or more) numbers is the smallest number that's a multiple of both numbers. In other words, it's the smallest number that the other numbers all "go into".

One way to find the LCM is by using prime factors. The LCM of a set of numbers has to contain all the primes of the individual numbers in the set.

For example, if our numbers are:

5, 6 and 9

then our LCM will have to contain 5, 2&3 (from 6) and 3&3 (from 9).

If we were to multiply all of those primes together, we'd get:

5 * 2 * 3 * 3 * 3 = 270

Now, 270 is a common multiple of 5, 6 and 9, but it's not the LOWEST common multiple.

When two (or more) numbers in the set have prime factors in common, we don't need to duplicate those primes to generate the LCM.

In our example, both 6 and 9 have prime factors of 3, so we eliminate one of the 3s to get our LCM of 90.

(Note that we still need 2 3s, since 9 has two prime factors of 3 and if we were to eliminate one of those 9 would no longer be a factor of our number.)
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by nadontheway » Tue Jan 29, 2008 2:32 pm
Got u Stuart! Now if we were looking for the GCF of 5,6 and 9, we will have: 5*2*3=30. Is that correct?

By the way, (since I am a non native english speaker), are "Greatest Common Factor" and "Greatest Common Divisor" equivalent? have the meaning?
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by Stuart@KaplanGMAT » Tue Jan 29, 2008 3:40 pm
nadontheway wrote:Got u Stuart! Now if we were looking for the GCF of 5,6 and 9, we will have: 5*2*3=30. Is that correct?

By the way, (since I am a non native english speaker), are "Greatest Common Factor" and "Greatest Common Divisor" equivalent? have the meaning?
Greatest common factor is indeed the same as greatest common divisor.

Since 5, 6 and 9 have no prime all in common, the GCF is 1. It's only if numbers have primes in common that the GCF will be greater than 1.

For example, let's find the GCF of 12, 21 and 30

12 = 2 * 2 * 3
21 = 3 * 7
30 = 2 * 3 * 5

Since the only prime factor that these 3 numbers have in common is "3", the GCF is 3.

If our 3 numbers were 12, 42 and 60:

12 = 2 * 3 * 3
42 = 2 * 3 * 7
60 = 2 * 2 * 3 * 5

These three numbers have "2" in common and "3" in common. Therefore, the GCF is 2 * 3 = 6

Hope that helps!
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by nadontheway » Tue Jan 29, 2008 4:10 pm
Thank you so much Stuart!! that makes my ideas clearer now.

One last question, if you don't mind. I read in a book that the GCF of
60= 2*2*3*5 and
72= 2*2*2*3*3

is 2*2*3=12

but as per your explanations it should only be 2*3=6. Am I right?
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by Stuart@KaplanGMAT » Tue Jan 29, 2008 7:22 pm
nadontheway wrote:Thank you so much Stuart!! that makes my ideas clearer now.

One last question, if you don't mind. I read in a book that the GCF of
60= 2*2*3*5 and
72= 2*2*2*3*3

is 2*2*3=12

but as per your explanations it should only be 2*3=6. Am I right?
The book is correct... 60 and 72 have two 2s and one 3 in common, so the GCF will be 2 * 2 * 3
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by nadontheway » Wed Jan 30, 2008 5:32 am
Ohh that could be tricky...but I guess I understood now. thx so much for your help Stuart.
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by richardwang6430 » Tue Apr 15, 2008 6:05 pm
thank you, stuart
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by pharmxanthan » Thu Aug 05, 2010 6:39 pm
Stuart Kovinsky wrote:If x is a positive integer, what's the LCM of x, 6 and 9?

(1) The LCM of x and 6 is 30.

We can attack this statement by brute force or by using our knowledge of principles.

If the LCM of x and 6 is 30, then x could be 5, 10, 15 or 30. If you try out all 4 of those possibilities, you'll get a LCM of x, 6 and 9 of 90. Since we get the same answer every time, (1) is sufficient.

Alternatively, if the LCM Of x and 6 is 30, that means that x must have a prime factor of 5 (since the "5" in 30 has to come from either x or 6, and it doesn't come from 6). So, we need a "5" from x, a "2" and "3" from 6 and another "3" from 9. 2*3*3*5 = 90. Sufficient.

(2) the LCM of x and 9 is 45.

Same solution as above!

The possibly values for x are 5, 15 and 45. In all 3 cases, the LCM of x, 6 and 9 is 90. Sufficient.

Alternatively, if the LCM of x and 9 is 45, once again the prime factor of "5" has to come from the x. We're going to end up with 2*3*3*5 yet again. Sufficient.

Both statements are sufficient alone: choose (D).
How to pick numbers if LCM and GCF are given? (see the bold statements)
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by chetanc » Wed Mar 30, 2011 12:28 am
Hello Stuart,

Can you explain the principle, i.e how can X be 5,10,15 or 30 and in the second explanation why do you take a 3 from 9 to get 90 .
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