Sum of Products of Subsets

We can re-interpret the sum of the f(A) for all A as

$\sum_{n = 0}^{100} (2^n)(g(n))$

Where g(n) = "the number of ways to partition n into the sum of any number of distinct non-negative integers less than or equal to 100"

It is important to say "non-negative" because we know $f(A) = (2^{a_1}) ( 2^{a_2} )...(2^{a_k})$

for some k positive integers a_i, and since 1 = 2^0 may be an element of A, one of the a's may be zero.

Daunting as this "g" function and its poorly worded definition may seem, it can be recognized as the coefficient of x^n in the expression $(x^0+1)(x^1+1)...(x^{100} + 1) - 1$

(The "-1" corrects for the case of an empty set A.)

Thus our answer will be y, as defined below:

$(2^0+1)(2^1+1)...(2^{100} + 1) - 1 = y (mod 126)$

Since 126 = 9x2x7, and 2 and 9 are factors of y+1, we must simply evaluate y + 1 mod 7. Since the order of 2 mod 7 is 3, our (2^n + 1) terms are all either 2, 3, or 5 mod 7, and thus we have $y+1 = [(2)(3)(5)]^{33}(2)(3) = -2^{33} = -1 (mod 7)$

we have that y +1 is a multiple of 18 congruent to 6 mod 7, and is thus 90.

So, finally, $y= \boxed{89}$

Let $e_k$ be the $k$ th degree elementary symmetric polynomial

Then the sum of all $f(A) = e_1 + e_2 + e_3 + ... + e_{100}$

$f(A) = e_1 + e_2 + e_3 + ... + e_{100}$

$f(A) = (e_1 + 1)(e_2 + 1)(e_3 + 1)...(e_{100} + 1) - 1$

Using modulo $126$ ,

$f(A)(mod 126) = 2(3)(5)(17)(33)(65)(3)(5)...(3)(5)(17)(33) - 1$

When simplified,

$f(A)(mod 126) = \boxed{89}(mod 126)$

Sorry, i need to hurry. At least, i gave you an idea/outline on solving this problem.