M28-39

Official Solution:


If \(x\) and \(y\) are both positive integers and \(x \gt y\), what the remainder when \(x\) is divided by \(y\)?

Given that \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(x = yq + r\) and \(0 \leq r < y\).

(1) \(y\) is a two-digit prime number.

This statement is clearly insufficient, as we have no information about \(x\).

(2) \(x=qy+9\), for some positive integer \(q\).

It might be tempting to consider this statement sufficient and conclude that \(r = 9\), given that the equation resembles \(x = yq + r\). However, we don't know whether \(y > 9\): the remainder must be less than the divisor.

For example:

If \(x = 10\) and \(y = 1\), then \(10 = 1*1 + 9\), but the remainder upon dividing 10 by 1 is zero.

If \(x = 11\) and \(y = 2\), then \(11 = 1*2 + 9\), but the remainder upon dividing 11 by 2 is one.

Not sufficient.

(1)+(2) From (2), we know that \(x = qy + 9\), and from (1), we know that \(y\) is greater than 9 (since it's a two-digit number). Thus, we have a direct formula for the remainder, as given above. Sufficient.


Answer: C