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If positive integer x..

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by Elena89 » Wed Dec 14, 2011 2:35 am
If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105 ?

(1) x is a multiple of 9
(2) y is a multiple of 25

Source: OG-12

[spoiler]OA: (B)[/spoiler]

Im unable to understand OG's explanation for this problem. Can someone explain how to solve such kind of DS problems in a simpler way?
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Source: — Data Sufficiency |

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by Anurag@Gurome » Wed Dec 14, 2011 2:42 am
Elena89 wrote:If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105 ?

(1) x is a multiple of 9
(2) y is a multiple of 25

Source: OG-12

[spoiler]OA: (B)[/spoiler]

Im unable to understand OG's explanation for this problem. Can someone explain how to solve such kind of DS problems in a simpler way?
x = 2 * 3
y = 2 * 7
105 = 3 * 5 * 7
x * y = 2² * 3 * 7
For xy to be a multiple of 105, we need at least one 3, 5, and 7 among the factors. So, the missing number is 5, as 3 and 7 are already there.

(1) x is a multiple of 9 but this does not imply anything about the missing factor, 5; NOT sufficient.

(2) y is a multiple of 25 and 25 = 5 * 5, which answers the question; SUFFICIENT.

The correct answer is B.
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by smiles@Gmat » Mon Aug 27, 2012 9:47 pm
Anurag ,

x = 2 * 3
y = 2 * 7
105 = 3 * 5 * 7
x * y = 2² * 3 * 7


My thought was X * Y can have 2 * 3 * 7 definitely (Where 2 in either X or Y can be redundant).
Please clarify
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by neelgandham » Mon Aug 27, 2012 10:53 pm
If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105 ?.

I will try and rephrase the question for better understanding.
Since, x is a multiple of 6 it can be written in the form x = 6*m, where m is a non zero integer.
Since, y is a multiple of 14 it can be written in the form y = 14*n, where y is a non zero integer.

x*y = 6*m*14*m = 2*3*7*2*m*n = (105)*(4*m*n/5). For x*y to be a multiple of 105, 4*m*n/5 should be an integer. i.e. m or n should be a multiple of 5. The question can be rephrased to

'If x = 6*m and y = 14*n, Is m or n a multiple of 5?'
(1) x is a multiple of 9
If x = 18 then the value of m is 3. m is not a multiple of 5.
If x = 90 then the value of m is 15. m is a multiple of 5.
So, statement I is insufficient to answer the question.
(2) y is a multiple of 25
y is a multiple of 25 and 14. So, y is a multiple of 25*14(because you con't have any common multiples between 25 and 14). i.e. y = 350*l, where l is a non zero integer.
Now, y = 350*l = 14*n.
25*l = n. So, n is a multiple of 25(and of 5).

Statement II is sufficient to answer the question.
IMO B
p.s: It is just another approach but Anurag's approach is the better one.
Anil Gandham
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