# Do you really like 11 so much?

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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