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

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

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?

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

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

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

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

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

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


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



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

Regards,
Fabio.

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