Inspired from: Problem 2 : Basic Operations by Ahaan Rungta
A real number is given. In each move, we can either add to it, subtract to it, multiply it by and divide it by . The sum of all numbers such that after exactly moves, the original number comes back is . Find .
The answer is 0.
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1 solution
For every real number x satisfying the problem, we call the moves that x follows f i ( x ) ; 1 ≤ i ≤ 7 . Let f i ′ ( x ) be the move that satisfy the conditions: f i ′ ( x ) ≡ f i ( x ) if f i ( x ) = x ∗ 7 or f i ( x ) = x / 7 ; f i ′ ( x ) = x − 7 if f i ( x ) = x + 7 and f i ′ ( x ) = x + 7 if f i ( x ) = x − 7 .
We prove that f i ( … f 1 ( x ) ) = − f i ′ ( … f 1 ′ ( − x ) )
It can easily be proven that f i ( a ) = − f i ′ ( − a ) (due to the nature of f i ′ ( x )
Therefore f i ( … f 1 ( x ) ) = − f i ′ ( … f 1 ′ ( − x ) )
Or f i ′ ( … f 1 ′ ( − x ) ) = − x
Because x + − x = 0 , therefore a = 0 , resulting in [ a ] = 0
Sorry I do not know how to present a solution in English properly nor am I any good in expressing my solution, so I hope that you can understand it