smc

First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. Say that $N\equiv a(\mathrm{mod}6)$ also that $N\equiv b(\mathrm{mod}5)$ Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b<5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$. $N\equiv a(\mathrm{mod}6)$, $N\equiv a(\mathrm{mod}5)$, $⟹N=a(\mathrm{mod}30)$, $0\le a\le 4$ Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$ Just keep in mind that $y$ can be one of five choices: $0,1,2,3,$ or $4$, ; Also, we have already found which digits of $y$ will add up into the units digits of $2N$. Now, examine the tens digit, $x$ by using $\mathrm{mod}25$ and $\mathrm{mod}36$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work) Then we take $N=30x+y$ $\mathrm{mod}25$ and $\mathrm{mod}36$ to find the last two digits in the base $5$ and $6$ representation. $$N\equiv 30x(\mathrm{mod}36)$$ $$N\equiv 30x\equiv 5x(\mathrm{mod}25)$$ Both of those must add up to $$2N\equiv 60x(\mathrm{mod}100)$$ ($33\ge x\ge 4$) Now, since $y=0,1,2,3,4$ will always work if $x$ works, then we can treat $x$ as a units digit instead of a tens digit in the respective bases and decrease the mods so that $x$ is now the units digit. $$N\equiv 5x(\mathrm{mod}6)$$ $$N\equiv 6x\equiv x(\mathrm{mod}5)$$ $$2N\equiv 6x(\mathrm{mod}10)$$ Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x) Then $$N\equiv 5m+n(\mathrm{mod}5)$$ $$N\equiv 25m+5n(\mathrm{mod}6)$$ $$2N\equiv 30m+6n(\mathrm{mod}10)$$ and this simplifies to $$N\equiv n(\mathrm{mod}5)$$ $$N\equiv m+5n(\mathrm{mod}6)$$ $$2N\equiv 6n(\mathrm{mod}10)$$ From careful inspection, this is true when $n=0,m=6$ $n=1,m=6$ $n=2,m=2$ $n=3,m=2$ $n=4,m=4$ This gives you $5$ choices for $x$, and $5$ choices for $y$, so the answer is $5\ast 5=\boxed{25}$