GMATPrepNow wrote:

W, X, Y and Z are four different positive integers. When X is divided by Y, the quotient is Z and the remainder is W.

What is the value of Z?

1) W = X – 4

2) W + Z = 4

Beautiful problem, Brent. Kudos!

\(W,X,Y,Z\,\,\, \ge 1\,\,\,{\rm{different}}\,\,{\rm{ints}}\,\,\left( * \right)\)

\(\left\{ \matrix{

X = ZY + W\,\,,\,\,\,W < Y\,\,\,\left( {**} \right) \hfill \cr

\left( * \right)\,\,\,1 \le W < Y\,\,\, \Rightarrow \,\,\,Y \ge 2\,\,\,\left( {***} \right) \hfill \cr} \right.\,\)

\(? = Z\)

\(\left( 1 \right)\,\,X = W + 4\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,W + 4 = ZY + W\,\,\,\, \Rightarrow \,\,\,\,\,ZY = 4\,\,\,\,\mathop \Rightarrow \limits_{\left( {***} \right)}^{\left( * \right)} \,\,\,\,\,\,\left( {Y,Z} \right) = \left( {4,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)

(The viability of the unique (Y,Z)=(4,1) is an examiner´s burden. In this case it is nice to mention, as cheetan2u did, that (X,Y,Z,W)=(7,4,1,3) is viable. The problem is perfect!)

\(\left( 2 \right)\,\,W + Z = 4\,\,\,\,\left\{ \matrix{

\,{\rm{Take}}\,\,\left( {X,Y,Z,W} \right) = \left( {7,2,3,1} \right)\,\,\,\, \Rightarrow \,\,\,? = 3\,\, \hfill \cr

\,{\rm{Take}}\,\,\left( {X,Y,Z,W} \right) = \left( {7,4,1,3} \right)\,\,\,\, \Rightarrow \,\,\,? = 1 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}\)

The correct answer is (A).

We follow the notations and rationale taught in the

GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik ::

GMATH method creator (Math for the GMAT)

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