If $b_nb_{n-1} \cdots b_0$ is the binary representation of natural $x$, and if $(b_0-b_1+b_2-b_3+\cdots+(-1)^nb_n)=0\pmod3$, then $x=0\pmod3$

Solution 1:

As for how to come up with a solution, sometimes testing with purposeful simple examples could give insights to these problems.

Observe:
$1_{(10)} = 2^0_{(10)} = 1_{(2)}$. The sum is $1 \pmod{3}$.
$2_{(10)} = 2^1_{(10)} = 10_{(2)}$. The sum is $-1 \pmod{3}$.
$4_{(10)} = 2^2_{(10)} = 100_{(2)}$. The sum is $1 \pmod{3}$.
$8_{(10)} = 2^3_{(10)} = 1000_{(2)}$. The sum is $-1 \pmod{3}$.

Observe that as the exponent of $2_{(10)}$ increases, the sum$\pmod{3}$ alternates between $1$ and $-1$.
Would it remind you of some patterns?

Also, notice that base $10$ is not mentioned in the question. This may mean it is irrelevant, and we probably have better chances if we construct our examples in base $2$.

Solution 2:

$$x = \sum_{i=0}^n b_i2^i$$

Notice that we have $2 \equiv -1 \pmod{3}$, can you complete the proof?