Number theory

Here is a lucky solution: Let $n$ be the number. Because we are told that both $n$ and $6n$ have 6 digits, it must be that $6a < 10$ . The only natural solution for this inequality is $a = 1$ . Furthermore, because the product $2n$ has the same digits as $n$ , the digit sum is the same so $2n \equiv n \mod 9 \implies n \equiv 0 \mod 9$ . And $s \equiv n \equiv 0 \mod 9$ .

But $s = 1 + b + c + d + e + f \leq 1 + 9 + 9 + 9 + 9 + 9 = 46$ this means that the only possible values for $s$ are $9, 18, 27, 36 , 45$ . Brilliant allows you to try 3 of those so you have a $60$ % chance of solving the problem by trying 3 of those. I tried $9,18, 27$ .

Recall that

$1÷7=0.\overline{142857}$

This repeating series of digits occurs after the decimal whenever a number that is not a multiple of 7 is divided by 7. For example:

$2÷7=0.2857\overline{142857}$

This means that the number $142857$ , when multiplied by a number between 1 and 7, will result in the same digits, except some of the digits in the front have been shifted to the back. Therefore $142857$ is a number that satisfies the problem.

$\times 1 = 142857$

$\times 2 = 285714$

$\times 3 = 428571$

$\times 4 = 571428$

$\times 5 = 714285$

$\times 6 = 857142$

$1+4+2+8+5+7= \boxed{27}$