If x and y are positive integers and x/y has a remainder of 5, what is

THEORY:
If \(a\) and \(d\) are positive integers, there exists unique integers \(q\) and \(r\), such that \(a = qd + r\) and \(0\leq{r}<d\). Where \(a\) is a dividend, \(d\) is a divisor, \(q\) is a quotient (\(\geq0\)) and \(r\) is a remainder.

Now consider the case when dividend \(a\) is less than divisor \(d\). For instance: what is the ramainder when \(a=7\) divided by \(d=9\):

\(a = qd + r\) --> \(7=q*9+r\), as \(q\geq0\) and \(0\leq{r}<d\) --> \(7=0*9+r=0*9+7\) --> \(r=remainder=7\).

Hence in ANY case when positive integer \(a\) is divided by positive integer \(d\) and given that \(a<d\), remainder will always be \(a\). Note here that EVERY GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

So: \(\frac{7}{9}\), remainder \(7\); \(\frac{11}{13}\), remainder \(11\); \(\frac{135}{136}\), remainder \(135\).



Back to our original question:
If x and y are positive integers and x/y has a remainder of 5, what is the smallest possible value of xy?

A. 6
B. 12
C. 18
D. 24
E. 30

When we got that \(x=5\) and \(\frac{5}{y}\) gives remainder of \(5\), according to the above we can conclude that \(y\) must be more than \(x=5\). As we want to minimize the \(x*y\) we need least integer \(y\) which is more than \(x=5\). Thus \(y=6\) and \(xy=30\).

Check it:
\(5=0*6+5\) --> \(x=5\) dividend, \(y=6\) divisor, \(q=0\) quotient and \(r=5\) remainder.

Answer: E.

Hope it helps.