jmc

This is true if and only if $f(n):=6+3n+2n^2+5n^3+3n^4+2n^5$ is a multiple of $7$. Whether or not this is true depends only on $n$ modulo $7$. First note that the polynomial is congruent to $2n^5+3n^4+5n^3+2n^2+3n-15$ modulo $7$, which has $1$ as a root. Factoring, we get $$2n^5+3n^4+5n^3+2n^2+3n-15=(n-1)(2n^4+5n^3+10n^2+12n+15).$$Next we check each residue modulo $7$, i.e. we check this for $n=2,3,-1,-2,-3$. Since $n-1$ is not a multiple of $7$ when $n$ is not congruent to $1$ modulo $7$, we need only check the quartic factor. When $n=2$, we get $2(16)+5(8)+10(4)+12(2)+15=32+40+40+24+15=112+39=151$, which is not a multiple of $7$. When $n=-1$, we get $15-12+10-5+2=10$, which is not a multiple of $7$. When $n=-2$, we get $$32-40+40-24+15=32+15-24=8+15=23,$$which is not a multiple of $7$. When $n=3$, we get $2(81)+5(27)+10(9)+12(3)+15=162+135+90+36+15=297+126+15=312+126=438$, which is again not a multiple of $7$. Finally, when $n=-3$, we get $162-135+90-36+15=338-2(135)-2(36)=438-270-72=168-72=96$, which is again not a multiple of $7$. Thus the only possible $n$ are those congruent to $1$ modulo $7$, and furthermore note that $n\ge 7$ since $6$ is a digit. Thus the possible values of $n$ are $7m+1$ for $1\le m\le 14$, so there are $\boxed{14}$ possible values.