Bunuel wrote:

Let b and x be positive integers. If b is the greatest divisor of x that is less than x, is the sum of the divisors of x, which are less than x itself and greater than one, greater than 2b?

(1) b^2 = x

(2) 2b = x

Statement 1: \(b\) is the greatest divisor of \(x\) and \(x=b^2\), so \(b\) must be prime.

For eg. \(b=7\), then \(x=49\). Divisors of \(x=1,7,49\). Greatest divisor \(7=b\)

\(b\) cannot be a composite number. for eg. if \(b=4\), then \(x=16\). Divisors of \(x= 1,2,4,8,16\). Greatest divisor \(8\) which is not equal to \(b\).

So if \(b =p_1\), a prime number and as \(x\) is square of a prime number, \(x={p_1}^2\), divisors of \(x\) will be \(1\), \(p_1\) & \({p_1}^2\) only.

Now the divisors of \(x\), which are less than \(x\) itself and greater than one is \(p_1=b\) which is not greater than \(2b=2p_1\).

SufficientStatement 2: if \(b=3\), \(x=6\). Divisors of \(x= 1,2,3,6\). sum of divisors of \(x\), which are less than \(x\) itself and greater than one is \(2+3=5<2*3\)

if \(b=6\), \(x=12\). Divisors of \(x=1,2,3,4,6,12\). sum of divisors of \(x\), which are less than \(x\) itself and greater than one is \(2+3+4+6=15>2*6\). Hence

InsufficientOption

A