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**Prove It** Ok, we want to see if the number $\displaystyle \displaystyle x_nx_{n-1}x_{n-2}\dots x_3x_2x_1$ is divisible by 3.

Since $\displaystyle \displaystyle x_1$ is in the units column, its value is $\displaystyle \displaystyle x_1 = 10^0x_1$.

Since $\displaystyle \displaystyle x_2$ is in the 10s column, its value is $\displaystyle \displaystyle 10x_2 = 10^1x_2$.

Since $\displaystyle \displaystyle x_3$ is in the 100s column, its value is $\displaystyle \displaystyle 100x_3 = 10^2x_3$.

$\displaystyle \displaystyle \vdots$

Continuing in this fashion we find that $\displaystyle \displaystyle x_n$ is in the $\displaystyle \displaystyle 10^{n-1}$s column, so its value is $\displaystyle \displaystyle 10^{n-1}x_n$.

Also note that every power of 10 will have one remainder when divided by 3 (since they can be written as 9 + 1, 99 + 1, 999 + 1, etc.

So the value of $\displaystyle \displaystyle x_nx_{n-1}x_{n-2}\dots x_3x_2x_1$ is

$\displaystyle \displaystyle \begin{align*}&\phantom{=}10^{n-1}x_n + 10^{n-2}x_{n-1} + 10^{n-3}x_{n-2} + \dots + 10^2x_3 + 10x_2 + x_1 \\ &= \left(10^{n-1} - 1\right)x_n + \left(10^{n-2}-1\right)x_{n-1} + \left(10^{n-3}-1\right)x_{n-2} + \dots + \left(10^2 - 1\right)x_3 + \left(10^1 - 1\right)x_2 + x_n + x_{n - 1} + x_{n-2} + \dots + x_3 + x_2 + x_1 \end{align*}$

Each of the $\displaystyle \displaystyle 10^k-1$ terms is divisible by 3, so the number will only be divisible by 3 if the sum of the remaining terms is divisible by 3.

So the number $\displaystyle \displaystyle x_nx_{n-1}x_{n-2}\dots x_3x_2x_1$ is divisible by 3 if $\displaystyle \displaystyle x_n + x_{n-1} + x_{n-2} + \dots + x_3 + x_2 + x_1$ is divisible by 3.