Alexey1989x wrote:
The sum of \(q\) and \(m\) equals to some positive odd number, where \(q\) and \(m\) are prime numbers.
When \(30\) is divided by \(q^k\) it leaves the remainder \(r\), where \(k\) is a positive integer. What is the value of \(r\) ?
(1) \(15 < q^k < 40\)
(2) \(q < m\) and \(q^k\) leaves \(2\) as a remainder when divided by \(7\)
Self-made
Analyzing question stem we can see that either q or m is 2 because q and m are prime numbers and their sum is odd. Since E+O = O and except 2 all other prime numbers are odd.
Now 30 = q^k*(Quotient) +r. this implies that q^k<30 because if q^k>=30 then either there will be no remainder (if q^k=30) or the remainder will be 30 itself.
Statement 1: if q=2 then from this statement q^k could be 16 or 32 i.e. powers of 2 but if q is not equal to 2 then it could be 27, 25. Hence the statement is not sufficient
Statement 2: As q<m this means q has to be 2 and m could be any prime no greater than 2. Also from this statement we get to know that q^k = 7n+2 (where "n" is the quotient). so we have 2^k = 7n+2
if n = 2; q^k = 16 and if n = 18; q^k = 128. Hence this statement is not sufficient.
Combining Statement 1 & 2 we get q^k = 16.
So option
C