A number is formed using the digits $2,4,7$ exactly once, and using as many zeroes as required for adjustment.

Then $S$ be the set of remainders which can **NOT** be obtained on dividing the number by $11$ . Find the sum of all elements of $S$

**Details and assumptions** :-

$\bullet\quad$ A set contains each element exactly once, no repetitions.

$\bullet \quad S$ only contains some integers $k_i$ in the range $0\leq k_i \leq 10$, as remainder is , by definition, $0\leq remainder < divisor$.

$\bullet \quad$ You are expected to find the sum of all elements in $S$, that is the remainders which can't be obtained by dividing the number by $11$.

$\bullet\quad$ The number $2470$ is a a valid number for consideration, and so is $204007$ and all others which contain only the digits $2,4,7,0$ and $2,4,7$ are exactly once. (Thus, $2247$ is not a valid number)

This is a part of set 11≡ awesome (mod remainders)

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