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what is the sum of the m and n?

by vdm » Mon Sep 14, 2015 6:55 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
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by GMATGuruNY » Mon Sep 14, 2015 6:56 pm
vdm 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
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 vdm » Mon Sep 14, 2015 9:42 pm
Thank you, Mitch! Your explanation made it real simple to understand.
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by Max@Math Revolution » Tue Sep 15, 2015 8:43 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.


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

Transforming the original condition and the question we have (m,n)=(56*3,56*5),(56*1,56*15),(56*15*56*1),(56*5,56*3). Finding out m automatically gives us n, therefore we have 2 variables and thus we need 2 equations to match the number of variables and equations. Since there is 1 each in 1) and 2), D is likely the answer.

In case of 1), (m,n)=(56*3,56*5),(56*1,56*15),(56*5,56*3) therefore the answer is not unique and the condition is not sufficient.
In case of 2), (m,n)=(56*1,56*15) is unique, therefore the condition is sufficient. Thus the answer is B.

Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.

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