Here's a nice trick that would work on this question, or any question involving the sum of the factors of a number; I wanted to share it because I think it's (i) cool and (ii) new to most students.

If we know the prime factorization of an integer, we can find the sum of the factors as follows:

1:: Find the highest power of each prime factor
2:: Find the sum of ALL the powers of that prime factor, up to the highest power present
3:: Multiply these sums together

For instance, if I want the sum of the factors of 360, I follow those steps like so::

1:: 360 = 2³ * 3² * 5
2:: The sums of the powers are (2� + 2¹ + 2² + 2³) and (3� + 3¹ + 3²) and (5� + 5¹).
3:: Multiplying those sums gives (2� + 2¹ + 2² + 2³) * (3� + 3¹ + 3²) * (5� + 5¹) = 15 * 13 * 6 = 1170.

Too cool!

Now let's look at our problem.

S1 tells us the sum is 31. 31 is prime, so it can only be represented as the product 1 * 31. This means that we can only have ONE prime factor, as we can't be multiplying two sums greater than 1.

Now let's see what sums of the first few powers of certain primes give us 31.

(2� + 2¹ + 2² + 2³ + 2�) = 31, so this is a possibility
(3� + 3¹ + 3² + 3³) = 40, so no power of 3 will work
(5� + 5¹ + 5²) = 31, so this is a possibility too

Hence we could have 2� or 5²; INSUFFICIENT.

S2 obviously doesn't do much on its own.

Taking the two together, we still have 16 and 25 as possibilities; INSUFFICIENT.