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If z is a multiple of 9 and w is a multiple of 4, is zw

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If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post Updated on: 27 Aug 2013, 01:36
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If z is a multiple of 9 and w is a multiple of 4, is zw a multiple of 126?

(1) z is a multiple of 21
(2) w is a multiple of 25

Originally posted by guerrero25 on 26 Aug 2013, 22:50.
Last edited by Bunuel on 27 Aug 2013, 01:36, edited 1 time in total.
Moved to DS forum and added the OA.
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 26 Aug 2013, 23:58
2
guerrero25 wrote:
If z is a multiple of 9 and w is a multiple of 4, is zw a multiple of 126?

A) z is a multiple of 21
B) w is a multiple of 25


IMO A

126 = 9 * 7 * 2

NOW given
z= has atleast one 9
w has atleast one 4
now for zw to be multiple of 126 we need (9 * 7 * 2) so we are lacking only 7

statement 1: z is a multiple of 21
therefore z= 7*3*k
hence we have 7 too in zw THEREFORE YES.
SUFFICIENT.

statement 2:w is a multiple of 25
means we have two 5 's for sure but no idea about 7.
hence this is not sufficient.

hence A
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 27 Aug 2013, 01:17
1
guerrero25 wrote:
If z is a multiple of 9 and w is a multiple of 4, is zw a multiple of 126?

A) z is a multiple of 21
B) w is a multiple of 25


From F.S 1, we know that z is a multiple of both 9 and 21. Thus, z will be of the form z = 63*q(q is an integer). Now, zw will be of the form 63*4*k(an integer) = (63*2)*2k = 126*2k. Thus, zw is a multiple of 126. Sufficient.

From F.S 2, for w=0,z=0, we know that zw is a multiple of 126. However, for w=100,z=9, zw is not a multiple of 126. Insufficient.

A.
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 27 Aug 2013, 01:47
2
Question :
z has at least two 3s, w has at least two 2s
Does zw have at least one 2, two 3s and one 7. i.e Do z or w have at least one 7 between them.

1) z has at least one 3 and one 7. Sufficient.
2) w has at least two 5s. Insufficient.

Answer is A
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 04 Nov 2014, 10:37
I don't get this answer; I thought that zero is a multiple of every number and hence z and w could both be zero leading to answer E? The question stem doesn't seem to exclude z and w from being zero.

What am I missing?
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 05 Nov 2014, 04:18
phferr1984 wrote:
I don't get this answer; I thought that zero is a multiple of every number and hence z and w could both be zero leading to answer E? The question stem doesn't seem to exclude z and w from being zero.

What am I missing?


But even if both z and w are 0, isn't zw = 0 a multiple of 126?
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 06 Jan 2015, 06:17
z is a multiple of 9 and w is a multiple of 4, is zw a multiple of 126?
Base work - zw = 126 = 3^2 * 2 * 7

A) z is a multiple of 21 and also 9 (given) so
z is a multiple of 63*3
w is a multiple of 4 hence zw has to be a multiple 126
A sufficient

B) w is a multiple of 25 and also 4 (given) so
w is a multiple of 100
z is a multiple of 9
In this case zw is a multiple of 900 (no 7 factor) and hence zw of case B is not a multiple of 126

Answer : A only sufficient
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Re: If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 13 Dec 2017, 10:19
guerrero25 wrote:
If z is a multiple of 9 and w is a multiple of 4, is zw a multiple of 126?

(1) z is a multiple of 21
(2) w is a multiple of 25


Lets look at the prime factorisation of 126. 126 = 2 * 3^2 * 7. So for a number to be a multiple of 126, that number must have at least one 2, two 3's and one 7. So we are given that z has at least two 3's (multiple of 9) and w has at least two 2's (multiple of 4). So zw already has the required 2's and 3's. All we need to know is whether zw also has a 7 or not?

(1) z is a multiple of 21 = 3*7. So z has a 7, thus we know zw has a 7 too. Sufficient.

(2) w is a multiple of 25 = 5^2. So zw has two 5's also, but we dont know about 7. Insufficient.

Hence A answer
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If z is a multiple of 9 and w is a multiple of 4, is zw  [#permalink]

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New post 24 Sep 2018, 04:27
guerrero25 wrote:
If z is a multiple of 9 and w is a multiple of 4, is zw a multiple of 126?

(1) z is a multiple of 21
(2) w is a multiple of 25

\(\frac{z}{{{3^2}}} = \operatorname{int} \,\,\,\,\,;\,\,\,\,\,\,\,\frac{w}{{{2^2}}} = \operatorname{int}\)

\(\frac{{zw}}{{2 \cdot {3^2} \cdot 7}}\,\,\,\mathop = \limits^? \,\,\,\operatorname{int}\)

\(\left( 1 \right)\,\,\,\,\left\{ \begin{gathered}
\,\frac{z}{{3 \cdot 7}} = \operatorname{int} \,\,\,\, \cap \,\,\,\,\frac{z}{{{3^2}}} = \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{z}{{{3^2} \cdot 7}} = \operatorname{int} \, \hfill \\
\,\frac{w}{{{2^2}}} = \operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\frac{w}{2} = \operatorname{int} \, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,\,\)

\(?\,\,\,:\,\,\,\,\frac{{zw}}{{2 \cdot {3^2} \cdot 7}} = \left( {\frac{z}{{{3^2} \cdot 7}}} \right) \cdot \left( {\frac{w}{2}} \right)\,\,\, = \,\,\,\operatorname{int} \,\, \cdot \,\,\,\operatorname{int} \,\,\,\, = \operatorname{int} \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


\(\left( 2 \right)\,\,\,\,\frac{w}{{{5^2}}} = \operatorname{int} \,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\,\left( {z,w} \right) = \left( {0,0} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,\,\left( {z,w} \right) = \left( {{3^2},{2^2} \cdot {5^2}} \right)\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{7}}\,\,{\text{missing}}} \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\,\,\,\,\,\)



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

Regards,
Fabio.
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