jmc

We let $N_7={\overline{a_n{a}_{n-1}\cdots a_0}}_7$; we are given that $$2(a_n{a}_{n-1}\cdots a_0{)}_7=(a_n{a}_{n-1}\cdots a_0{)}_{10}$$(This is because the digits in $N$ ' s base 7 representation make a number with the same digits in base 10 when multiplied by 2) Expanding, we find that $$2\cdot 7^n{a}_n+2\cdot 7^{n-1}{a}_{n-1}+\cdots +2a_0=10^n{a}_n+10^{n-1}{a}_{n-1}+\cdots +a_0$$ or re-arranging, $$a_0+4a_1=2a_2+314a_3+\cdots +(10^n-2\cdot 7^n)a_n$$ Since the $a_i$s are base-$7$ digits, it follows that $a_i<7$, and the LHS is less than or equal to $30$. Hence our number can have at most $3$ digits in base-$7$. Letting $a_2=6$, we find that $630_7={\boxed{315}}_{10}$ is our largest 7-10 double.