Do you really like 11 so much?

Number Theory Level 4

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)


Problem Loading...

Note Loading...

Set Loading...