Question 61
Find the number of $2013$-digit numbers $d_1 d_2 \dots d_{2013}$ with odd digits $d_1, d_2, \dots, d_{2013}$ so that
$d_1 \cdot d_2 + d_3 \cdot d_4 + \dots + d_{1809} \cdot d_{1810} \equiv 1 \pmod{4}$, $d_{1810} \cdot d_{1811} + d_{1811} \cdot d_{1812} + \dots + d_{2012} \cdot d_{2013} \equiv 1 \pmod{4}$.
$d_1 \cdot d_2 + d_3 \cdot d_4 + \dots + d_{1809} \cdot d_{1810} \equiv 1 \pmod{4}$, $d_{1810} \cdot d_{1811} + d_{1811} \cdot d_{1812} + \dots + d_{2012} \cdot d_{2013} \equiv 1 \pmod{4}$.