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GCF and LCM

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GCF and LCM

by manik11 » Mon Feb 29, 2016 5:02 am
If the greatest common factor of two integers, m and n, is 56 and the least common multiple is 840, what is the sum of the m and n?

(1) m is not divisible by 15.
(2) n is divisible by 15

OA : B

Source : Bellcurves
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by GMATGuruNY » Mon Feb 29, 2016 5:07 am
If the greatest common factor of two integers, m and n, is 56 and the least common multiple is 840, what is the sum of the m and n?

(1) m is not divisible by 15.
(2) n is divisible by 15
The GCF of m and n is composed of all of the prime factors that divide into both m and n.
56 = 2*2*2*7.
Thus, 2*2*2*7 divides into both m and n.

The LCM of m and n is composed of all of the prime factors that divide into only m, into only n, or into both m and n.
840 = 2*2*2*3*5*7.
Since 2*2*2*7 divides into both m and n, 3 and 5 each divide into only m or into only n.

Use Venn Diagrams to organize the data.
The following four cases are possible:
Image

Statement 1: m is not divisible by 15
Here, Cases 2, 3 and 4 are possible.
In Case 2, m = 2*2*2*7 and n= 2*2*2*3*5*7.
In Case 3, m = 2*2*2*3*7 and n = 2*2*2*5*7.
Since the value of m+n will be different in each case, INSUFFICIENT.

Statement 2: n is divisible by 15
Here, only Case 2 is possible.
In Case 2, m = 2*2*2*7 and n= 2*2*2*3*5*7.
Since the value of m+n can be determined, SUFFICIENT.

The correct answer is B.
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by DavidG@VeritasPrep » Mon Feb 29, 2016 5:12 am
manik11 wrote:If the greatest common factor of two integers, m and n, is 56 and the least common multiple is 840, what is the sum of the m and n?

(1) m is not divisible by 15.
(2) n is divisible by 15

OA : B

Source : Bellcurves
The GCF is 56 = 2^3 * 7. So we know that m and n each contain at least three 2's and one 7.

The LCM is 840 = 2^3 * 3 * 5 * 7 So we know that m or n (but not both) contain one 3 and that m or n (but not both) contain one 7.

S1: Test cases. We know that m does not contain both a 3 and a 5. Given our initial framework, we could have

Case 1: n= 2^3 * 3 * 5 * 7 and m= 2^3 * 7
Case 2 n= 2^3 * 5 * 7 and m= 2^3 * 3 * 7

These two cases will give different sums. S1 is not sufficient.

S2 Now we know for a fact that n contains one 3 and one 5. (And m cannot contain either. Otherwise, the GCF would not be 56.) The only scenario that will satisfy the parameters of the problem is

Case 1: n= 2^3 * 3 * 5 * 7 and m= 2^3 * 7 Because there is only one possible scenario, we can find m + n. Sufficient. Answer is B
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by Matt@VeritasPrep » Fri Mar 04, 2016 3:52 pm
I have a quick way that might be a little thorny at first, but once you get the idea, I think you'll like it.

Let's start with a great formula that work for any two positive integers x and y:

(GCF of x and y) * (LCM of x and y) = x * y

Applying that to our problem, we have

56 * 840 = m * n

We also know that both m and n are divisible by 56, so we can write both of them as 56*something, where "something" is a positive integer. Let's call these "somethings" a and b, respectively, so we have

m = 56*a
n = 56*b

Now we have

56 * 840 = 56a * 56b

Dividing both sides by 56, we have

840 = a * 56b

Dividing by 56 again, we have

15 = a * b

So we have two possibilities: 3 * 5 or 1 * 15. (In other words, either m = 56*3 and n = 56*5, or m = 56*1 and n = 56*15, or vice versa.) If we can solve for a + b, we're done, since we can easily plug these back into m + n = 56a + 56b. So the question becomes "What's a + b?"

S1::

m is not divisible by 15. So a = 1, 3, or 5. This ISN'T sufficient, since we could have a = 1, b = 15, or a = 3, b = 5, or vice versa, giving us either a + b = 16 or a + b = 8.

S2::

n is divisible by 15. So b must = 15, forcing a = 1, and leaving us with n = 56*15 and m = 56*1. Success!
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