`a)`
`S = 1 + 2 + 2^2 + 2^3 + ... + 2^2013 + 2014`
`2S = 2 + 2^2 + 2^2 + ... + 2^2013 + 2014 . 2`
`2S - S = S = 2014 - 1`
`S = 2013`
`=> S` hay `2013 : 3 = 671 (dư 0)`
Vậy `S : 3` dư `0`
`b)`
`(x + 1) + (x + 2) + ... + (x + 100) = 5750`
`(x + x + ... + x) + (1 + 2 + ... + 100) = 5750`
`100x + (1 + 100) . 100 : 2 = 5750`
`100x + 5050 = 5750`
`100x = 5750 - 5050`
`100x = 700`
`x = 700 : 100`
`x = 7`
Vậy `x = 7`