Bunuel wrote:
Official Solution:
If \(x\) and \(y\) are positive integers, there exist unique integers \(q\) and \(r\), called the quotient and remainder, respectively, such that \(x =divisor*quotient+remainder= yq + r\) and \(0 \leq r \lt y\).
(1) \(y\) is a two-digit prime number. Clearly insufficient since we know nothinf about \(x\).
(2) \(x=qy+9\), for some positive integer \(q\). It's tempting to say that this statement is sufficient and \(r=9\), since given equation is very similar to \(x =divisor*quotient+remainder= yq + r\) . But we don't know whether \(y \gt 9\)?: remainder must be less than divisor.
For example:
If \(x=10\) and \(y=1\) then \(10=1*1+9\), then the remainder upon division 10 by 1 is zero.
If \(x=11\) and \(y=2\) then \(11=1*2+9\), then the remainder upon division 11 by 2 is one.
Not sufficient.
(1)+(2) From (2) we have that \(x=qy+9\) and from (1) that \(y\) is more than 9 (since it's a two-digit number), so we have direct formula of remainder, as given above. Sufficient.
Answer: C
Hi ,
I'm still not able to understand this.
What will be dividend in this then ? If we are taking C both the statements ???