Unitary Divisors

Since $2016 = 2^5 \times 3^2 \times 7$ , we have $2016^{2016} = 2^{10080} \times 3^{4032} \times 7^{2016}$ . Thus the unitary divisors are $1, \hspace{0.5cm} 2^{10080},\hspace{0.5cm} 3^{4032}, \hspace{0.5cm} 7^{2016}, \hspace{0.5cm} 2^{10080}\times3^{4032}, \hspace{0.5cm} 2^{10080} \times 7^{2016}, \hspace{0.5cm} 3^{4032} \times 7^{2016}, \hspace{0.5cm} 2^{10080}\times3^{4032}\times7^{2016}$ and hence $S \; = \; \big(1 + 2^{10080}\big)\big(1 + 3^{4032}\big)\big(1 + 7^{2016}\big)$ We have $2^3 \equiv 0 \pmod{8}$ and $3^2 \equiv 7^2 \equiv 1 \pmod{8}$ , so that $2^{10080} \equiv 0 \pmod{8}$ and $3^{4032} \equiv 7^{2016} \equiv 1 \pmod{8}$ . Since $2^{100} \equiv 3^{100} \equiv 7^{20} \equiv 1 \pmod{125}$ , we deduce that $2^{10080} \equiv 2^{80} \equiv 51 \hspace{1cm} 3^{4032} \equiv 3^{32} \equiv 91 \hspace{1cm} 7^{2016} \equiv 7^{16} \equiv 101 \hspace{2cm} \pmod{125}$ By the Chinese Remainder Theorem, $2^{10080} \equiv 176 \hspace{1cm} 3^{4032} \equiv 841 \hspace{1cm} 7^{2016} \equiv 601 \hspace{2cm} \pmod{1000}$ and hence $S \equiv 177 \times 842 \times 602 \equiv \boxed{468} \hspace{1cm} \pmod{1000}$