The 74th question: The set $\mathrm{X}=\left\{1,2,3, \ldots, 5 \times 10^{6}\right\}$ has how many 2015-element subsets $A$ such that the sum of the elements of any non-empty subset of $A$ is not a multiple of 2016.
The 74th problem: How many 2015-element subsets \( A \) of the set \( X = \{1, 2, 3, \ldots, 5 \times 10^6\} \) satisfy the condition that the sum of the elements of any non-empty subset of \( A \) is not a multiple of 2016? (2016 Chen Shengshen Cup Mathematical Olympiad Problem)

Solution: We first prove a lemma: The sum of the elements of any non-empty subset of the integer set \( \{a_1, a_2, \ldots, a_{2015}\} \) is not a multiple of 2016 if and only if \( a_1 \equiv a_2 \equiv \ldots \equiv a_{2015} \equiv r \pmod{2016} \), where \( (r, 2016) = 1 \).

Proof of the lemma: According to the condition, the set \( \{0, a_1, a_1 + a_2, a_1 + a_2 + a_3, \ldots, a_1 + a_2 + a_3 + \ldots + a_{2015}\} \) forms a complete residue system modulo 2016. Similarly, the set \( \{0, a_2, a_2 + a_1, a_2 + a_1 + a_3, \ldots, a_2 + a_1 + a_3 + \ldots + a_{2015}\} \) also forms a complete residue system modulo 2016. Therefore, \( a_1 \equiv a_2 \pmod{2016} \). By the same reasoning, \( a_1 \equiv a_2 \equiv \ldots \equiv a_{2015} \equiv r \pmod{2016} \).

If \( (r, 2016) = d > 1 \), then \( \frac{2016}{d} < 2015 \), and \( a_1 + a_2 + \ldots + \frac{a_{2016}}{d} \equiv r \cdot \frac{2016}{d} \equiv 2016 \cdot \frac{r}{d} \equiv 0 \pmod{2016} \), which contradicts the condition. Hence, \( (r, 2016) = 1 \).

Conversely, if \( a_1 \equiv a_2 \equiv \ldots \equiv a_{2015} \equiv r \pmod{2016} \) and \( (r, 2016) = 1 \), then the sum of the elements of any \( k \) (where \( 1 \leq k \leq 2015 \)) elements of \( \{a_1, a_2, \ldots, a_{2015}\} \) is congruent to \( k \cdot r \pmod{2016} \), which clearly is not divisible by 2016.
Thus, the lemma is proved!

Now, let's return to the original problem.
Since \( 2016 = 2^5 \times 3^2 \times 7 \), we have \( \varphi(2016) = 2016 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 - \frac{1}{7}\right) = 576 \). Note that \( 5 \times 10^6 = 2016 \times 2480 + 320 \), and by the principle of inclusion-exclusion, the set \( \{1, 2, \ldots, 320\} \) contains exactly \( 320 - \left[\frac{320}{2}\right] - \left[\frac{320}{3}\right] - \left[\frac{320}{7}\right] + \left[\frac{320}{2 \times 3}\right] + \left[\frac{320}{2 \times 7}\right] + \left[\frac{320}{3 \times 7}\right] - \left[\frac{320}{2 \times 3 \times 7}\right] = 92 \) numbers that are coprime with 2016. Therefore, the number of 2015-element subsets \( A \) that satisfy the given condition is \( 92 \cdot \binom{2481}{2015} + (576 - 92) \cdot \binom{2480}{2015} = 92 \cdot \binom{2481}{2015} + 484 \cdot \binom{2480}{2015} \).