Pattern spotting

$\begin{aligned} \overline{bdc bdc} &= 100000b + 10000d + 1000c + 100b + 10d + c \\ \\ &= 100100b + 10010d + 1001c \\ \\ &= 1001(100b) + 1001(10d) + 1001(c) \\ \\ &= {\color{#D61F06}{1001}}\cdot(100b + 10d + c) \end{aligned}$

Let's start with this.

1. We don't know which letter represents which digit

2. We have to make a 6-digit number.

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Now, let number be uvwxyz(it's not multiplication) where u, v, w, x, y and z represent digits.

It can be written as

$10^5u + 10^4v + 10^3w + 10^2x + 10y + z$

= $(10^5u + 10^2x) + (10^4v + 10y) + (10^3w + z)$

Now, if (u, x), (v, y) and (w, z) are equal, we get a number divisible by 1001 as above number can be written as

$(100100u) + (10010v) + (1001w)$ = uvwuvw (again not multiplication)

Now, $4^{th}$ option corresponds to this value.

For a number to be divisible by $11$ , the sums of the alternate digits (e. g. the first, third, fifth,... digits and the second, fourth, sixth... digits) must be equal or differ by multiples of $11$ . Of the given options, only the fourth satisfy this condition.