factors

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factors

by MaxPower » Thu Dec 03, 2009 5:33 am
If K is a multiple of 29, is KY a multiple of 174?

(1) Y has all the same factors as 27.

(2) K is divisible by 2 without a remainder.
Source: — Data Sufficiency |

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by BuckeyeT » Thu Dec 03, 2009 5:57 am
C. Together, the answers are sufficient.

This can be solved using prime factorization. K is a multiple of 29. So, they share prime factors (29). Remember, K could have additional prime factors. But, it has at LEAST the same prime factors as 29.

We're trying to determine if KY is a multiple of 174. So, does KY share at least the same prime factors with 174 (2x3x29). We already know that K has 29 as a prime factor, so K and/or Y must have 2 and 3 as prime factors to be sufficient.

(1) Y shares factors with 27. So, they will share prime factors (3x3x3). This gives us one of the required values but not the other. Insufficient.

(2) K is divisivle by 2 without a remainder. So, K/2 = integer. K = 2 x integer. So, 2 is a prime factor of K. This gives us one of the required values but not the other.

(1) + (2) Sufficient. Together, we have the two required prime factors (2, 3).

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by Gmat09_5ALL » Tue Dec 08, 2009 11:27 am
IMO - A

statement 1 says that Y has all the same factor as 27 i.e. only factors of Y are (3*3*3) . Hence no 2 available .
Therefore NO

Hence A

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by mehravikas » Tue Dec 08, 2009 3:52 pm
Factors of 27 would be 1, 3 and 9

therefore A is insufficient.
Gmat09_5ALL wrote:IMO - A

statement 1 says that Y has all the same factor as 27 i.e. only factors of Y are (3*3*3) . Hence no 2 available .
Therefore NO

Hence A