2013 AMC 10B Problem 25

It is enough to work modulo $900=\operatorname{lcm}(25,36,100)$, because the last two base-$5$, base-$6$, and decimal digits repeat with that period. Let the last two base-$5$ digits be $a_1a_0$ and the last two base-$6$ digits be $b_1b_0$. The units digit condition gives $a_0+b_0\equiv2N\equiv2a_0\pmod{10}$, so $a_0=b_0$. Writing $N=150N_3+30a_1+a_0$, the base-$6$ tens digit condition gives $N_3\equiv a_1+b_1\pmod6$, so $N=900N_4+180a_1$ $+150b_1+a_0$. The tens digit condition for $S$ and $2N$ reduces to $5a_1\equiv b_1\pmod{10}$. With $0\le a_1\le4$ and $0\le b_1\le5$, the valid pairs are $(0,0),(2,0),(4,0),(1,5),(3,5)$. There are $5$ choices for $a_0=b_0$, so there are $5\cdot5=25$ choices of $N$. Thus, the correct answer is E .