Given that b is the greatest divisor of x, and 1<b<x, we are to determine if the sum of all the divisors (factors) of b greater than 1 and less than b is greater than 2b.
Statement 1: b^2=x
This implies that b=√x hence x must be a perfect square. From the given condition, possible values of x are:16, 36, 64, 81, 100, 144, 196, 225, 256, etc
Lets test with the smallest possible value of x=16
b=4.
factors of b that satisfy the condition =2. but 2b=8, since 2 < 8, the answer No.
Let x=20,712, the b=144
Factors of b satisfying given condition = 2,3,4,6,8,9,12,16,18,24,36,48,72 and their sum is 258 < 2b=288. The answer is therefore No to the question asked.
Statement 1 on its own is sufficient.
Statement 2: 2b=x, hence b=x/2
values of x that satisfy the given condition are: 8,12,18,20,24,...
when x=8, b=4, sum of factors of 4 satisfying the condition is 2, and 2<8, hence No.
x=24, b=12, factors of 12 satisfying the condition = 2,3,4,6 sum of the factors = 15. Since 15<24, the answer is still No.
Statement 2 is also sufficient on its own.
The answer is D.