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m is a multiple of 13. Is mn a multiple of 195?

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m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 07 Sep 2015, 03:45
1
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A
B
C
D
E

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Re: m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 07 Sep 2015, 05:23
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Bunuel wrote:
m is a multiple of 13. Is mn a multiple of 195?

(1) n has every factor that 45 has.

(2) m is divisible by 18.

Kudos for a correct solution.



the correct answer is A.

M is a multiple of 13. Therefore, the prime factorization of m atleast contains 13. ------- x

Is mn a multiple of 195? Rephrased question : Does the prime factorization of m*n atleast contain prime factors of 195, which are 3 5 and 13 ?

So basically we need to find out if prime factorization of m * n has atleast one 3 and one 5 as m*n already contains one 13 (given......x)

Statement 1) n has every factor that 45 has. the factors of 45 are 3,3 and 5. So, we can conclude that n has atleast one 3 and one 5.

Therefore, statement 1 is sufficient.

Statement 2) m is divisible by 18. i.e the prime factors of m atleast have 2, 3 and 3 as its prime factors.

But, we are not sure if m also has 5 as its factor. So statement 2 is insufficient.

Therefore, the correct answer is A.
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Re: m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 07 Sep 2015, 06:35
Solution: 195 = 13*3*5
m has 13 as a factor. For mn to be a multiple of 195, it should have 13,3 and 5 as factors. We know that m has 13. So, if we can prove that mn has 3 and 5 as factors, then mn will be multiple of 195.

Statement1 : n has same factors as 45. 45 has 3 and 5 as factors. So, mn has 3 and 5 as factors.
So, sufficient.

Statement2 : m has 18 as factor. So, it has 3 as a factor. We dont know whether mn has 5 as a factor or not. So, insufficient.

Option A.
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Re: m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 07 Sep 2015, 19:36
Prime factors of 195 = 3 x 5 x 13
m is a multiple of 13 ---> 13 is a prime factor of m
From (1): known factors of n = 3 x 3 x 5
--> known factors of mn = 3x3x5x13 --> minimum prime factors of 195 --> Sufficient
From (2): known factors of m = 2x3x3x13
Since no information about 5 --> Insufficient
Answer: A
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m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 08 Sep 2015, 06:41
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 equations ensures a solution.



m is a multiple of 13. Is mn a multiple of 195?

(1) n has every factor that 45 has.

(2) m is divisible by 18.


Transforming the original condition and the question, if mn=multiple of 195?--> mn=multiple of 3*5*13? and thus m is a multiple of 13. we need to know if n is has 3 and 5 as a factor. we have 2 variables (m,n) and 1 equation (m=multiple of 13) therefore we need 1 more equation, and D is likely the answer.
In case of 1), 45=3*3*5 therefore there is always 3 and 5 as a factor. Therefore the condition is sufficient.
In case of 2), there is no mention n therefore the condition is not sufficient. Therefore A is the answer.
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Re: m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 14 Sep 2015, 05:59
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Bunuel wrote:
m is a multiple of 13. Is mn a multiple of 195?

(1) n has every factor that 45 has.

(2) m is divisible by 18.

Kudos for a correct solution.


PRINCETON REVIEW OFFICIAL SOLUTION:

First, let’s make sure we are clear on what we’re looking for. This is a Yes/No Data Sufficiency problem, so it doesn’t matter whether mn is a multiple of 195 or not. That’s not what determines sufficiency or insufficiency. The key is whether we know for certain one way or the other. If we’re certain mn is a multiple of 195, that’s sufficient. If we’re certain mn is not a multiple of 195, that’s also sufficient. The data is insufficient only if we can’t tell.

We know m is a multiple of 13, but there’s not much to do with that, so let’s focus on the other number in the problem. The question wants to know whether mn is a multiple of 195, which is another way of asking whether 195 is a factor of mn, which is another way of asking whether mn is divisible by 195. So how do we figure out whether a number is divisible by 195 or not? By breaking it down into prime factors. 195 ends with a 5, so it must be divisible by 5, and we can start there. 195 ÷ 5 = 39. We can break down 39 a little more. 39 is 3 × 13. So the prime factorization of 195 is 5 × 3 × 13. What this means is that any number that is divisible by 5 and by 3 and by 13 is thereby divisible by 195. So returning to the question, we now have a way of figuring out whether mn is a multiple of 195. If mn is divisible by 5, 3, and 13, then it will be a multiple of 195. If mn is lacking any of those factors, then it won’t be. The question actually becomes even a little simpler, because we already know that m is a multiple of 13 — that’s given to us at the start. So the 13 is taken care of, and when we turn to the two statements we only need to look for the 3 and the 5. Let’s do that.

1) n has every factor that 45 has.

Does this tell us for certain whether mn is divisible by 3 and 5? Since 45 is divisible by 3 and 5, then n must be, and so must mn. Thus, Statement 1 is sufficient and we can eliminate answer choices B, C, and E.

2) m is divisible by 18.

Does this tell us for certain whether mn is divisible by 3 and 5? Well 18 is divisible by 3, but not by 5. So although we know that we have the 3 we’re looking for, we don’t know anything about the 5. It’s important to recognize that Statement 2 has not established that mn lacks 5 as a factor. We just don’t know. There’s nothing that prevents m from being divisible by 18 and also divisible by 5. Or n could be divisible by 5. Because we are ignorant about the 5, Statement 2 is insufficient, and the correct answer is A.
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Re: m is a multiple of 13. Is mn a multiple of 195?  [#permalink]

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New post 13 Jan 2019, 06:27
Bunuel wrote:
m is a multiple of 13. Is mn a multiple of 195?

(1) n has every factor that 45 has.

(2) m is divisible by 18.

\(m = 13K,\,\,K\,\,\operatorname{int} \,\,\,\,\,\left( * \right)\)

\(\frac{{m \cdot n}}{{3 \cdot 5 \cdot 13}}\,\,\mathop = \limits^? \,\,\operatorname{int} \,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\boxed{\,\,\frac{{m \cdot n}}{{3 \cdot 5}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,}\)


\(\left( 1 \right)\,\,\,n\,\,{\text{has}}\,\,{\text{5}}\,\,{\text{and}}\,\,{\text{3}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,\,{m \over {2 \cdot {3^2}}} = {\mathop{\rm int}} \,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {0,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2 \cdot {3^2} \cdot 13,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: m is a multiple of 13. Is mn a multiple of 195? &nbs [#permalink] 13 Jan 2019, 06:27
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