Too high to be counted
Let . Create every possible subset of and then make the product of the elements contained in every subset. How much is the sum of all the values you have obtained?
Submit your answer as the last 4 digits of this sum.
The answer is 9999.
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1 solution
Let p ( x ) be the polynomial of 2017th degree which zeros are − 2 0 1 7 , − 2 0 1 6 , − 2 0 1 5 , . . . , − 3 , − 2 , − 1 . Therefore p ( x ) can be written as: p ( x ) = ( x + 1 ) ( x + 2 ) ( x + 3 ) . . . ( x + 2 0 1 6 ) ( x + 2 0 1 7 ) By Vieta's formulas, every coefficient of x n , where n = 0 , 1 , 2 , 3 , . . . , 2 0 1 6 , is given by the sum of all possible products made by 2 0 1 7 − n zeros. This means that the sum we need is the sum of all the coefficient of p ( x ) , except for the coefficient of x 2 0 1 7 which is not made by any zero, as he can be created by taking always x from the brackets. The sum of the coefficients is p ( 1 ) = 2 ∗ 3 ∗ 4 ∗ 5 ∗ . . . ∗ 2 0 1 7 ∗ 2 0 1 8 which ends with 0000 as it cointains at least a 5 4 factor and at least a 2 4 factor. By subtracting the extra-coefficient of x 2 0 1 7 (which is 1), we obtains that the sum desidered ends with 9999.