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.
OA A
Source: Princeton Review
m is a multiple of 13. Is mn a multiple of 195?
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$$m = 13K,\,\,K\,\,{\mathop{\rm int}} \,\,\,\,\,\left( * \right)$$BTGmoderatorDC 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.
Source: Princeton Review
$$\frac{{m \cdot n}}{{3 \cdot 5 \cdot 13}}\,\,\mathop = \limits^? \,\,\operatorname{int} \,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\boxed{\,\,\frac{{m \cdot n}}{{3 \cdot 5}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int} \,\,}$$
$$\left( 1 \right)\,\,\,n\,\,{\rm{has}}\,\,{\rm{3}}\,\,{\rm{and}}\,\,{\rm{5}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,\,{m \over {2 \cdot {3^2}}} = {\mathop{\rm int}} \,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {0,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {m,n} \right) = \left( {2 \cdot {3^2} \cdot 13,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
This solution follows the notations and rationale taught in the GMATH method.
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Fabio.
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Given that m is a multiple of 13, we have m = 13k, where k is any integer.BTGmoderatorDC 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.
OA A
Source: Princeton Review
If mn a multiple of 195, mn = 195p, where p is any integer.
mn = 195p => 13kn = 195p => n = 195p/13k = 15p/k.
=> n = 15p/k
Let's take each statement one by one.
(1) n has every factor that 45 has.
Since 15 is a factor 45, the answer to the question is Yes. Sufficient.
(2) m is divisible by 18.
=> m is divisible by 13*18.
Thus, mn = 195p
13*18k*n = 195p
n = (5/6)*(p/k)
If p/k is a multiple of 6, the answer is Yes, else no. No unique answer. Insufficient.
The correct answer: A
Hope this helps!
-Jay
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