Question

1!+2!+....+50! is devided by 5! then what is a temainder

Answer 1

$Let \: the \: remainder \: be \: x \: \\ \\ 1! \: + \: 2! \: + \: 3! \: + \: ... \: +50! \: = x \: ( \: mod \: 5 \: ) \\ \\ 1! \: + \: 2! \: + \: 3! \: + 4! \: + \: 5! \: + \: 6 \times 5! + 7 \times 6 \times 5! \: + \: ... \:50 ! = \: x \: ( \: mod \: 5 \: ) \\ \\ Now \: , \: all \: terms \: greater \: than \: 4! \: \\ are \: divisible \: by \: 5 \: as \: they've \: 5 \: as \: \\ one \: of \: their \: factors \\ \\ \\ Hence \: , \: expression \: simplifies \: to \: ... \\ \\ 1! \: + \: 2! \: + \: 3! \: + 4! \: + \: 0 + 0 + 0 \: ...\: = x \: ( \: mod \: 5 \: ) \\ \\ 1 + 2 + 6 + 24 = 33 = x \: ( \: mod \: 5 \: ) \\ \\ \huge \: \boxed {x = 3}$

Answer 2

Answer:

heya mate. ur an in this attachment