manik11 wrote:If n is an integer, is the sum of all of n's divisors, which are not equal to n or equal to one, greater than 100?

(1) 3 and 96 are divisors of n that are neither equal to n nor equal to one.
(2) 3 and 97 are divisors of n that are neither equal to n nor equal to one.

OA : A
Source : BellCurves

Statement 1:
Any integer divisible by 96 will also be divisible by 3.
Since 96 is a factor of n that is not EQUAL to n, n must be a multiple of 96 such that n > 96:
192, 288...
If n is equal to any value in this list, its factors will include 2, 4 and 96, so the sum of its factors between 1 and n will be GREATER THAN 100.
SUFFICIENT.

Statement 2:
Since 3 and 97 are both prime, n must be a multiple of 3 and 97:
291, 582...
If n=291, then the sum of its factors between 1 and n = 3+97 = 100.
In this case, the answer to the question stem is NO.
If n=582, then its factors include 2, 3 and 97, so the sum of its factors between 1 and n will be GREATER THAN 100.
In this case, the answer to the question stem is YES.
INSUFFICIENT.

The correct answer is A.

We can also provide a neat formula for this.

Suppose we have the number 30. We know that 30 = 2 * 3 * 5, so any factor of 30 must have zero or one 2's, zero or one 3's, or zero or one 5's. To find ALL the factors, we just foil!

(1 + 2) * (1 + 3) * (1 + 5) =

(1 + 2 + 3 + 6) * (1 + 5) =

(1 + 2 + 3 + 6 + 5 + 10 + 15 + 30)

Aha!

So given any number, we can use a similar trick.

S1::

Since n > 96, we know n is a multiple of 96, e.g. 192, 288, etc. In any of those cases, our sum will be at least (1 + 3 + ...) * (1 + 2 + 4 + 8 + 16 + 32 + ...) - 1 - n. Since we have AT LEAST

(1 + 3) * (1 + 2 + 4 + 8 + 16 + 32) - 1 - 96

or

155, we know that we'll always have a sum > 100. Sufficient!

S2::

Our number could be 3*97, which gives a sum of (1 + 3) * (1 + 97) - 1 - 291, or 100, or it could be something much bigger, which would have a greater sum. Not sufficient!