what is the remainder when 1!+2!+3!+4!.....+2019! is divided by 21
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Answer:
l = a + (n - 1)d
2019! = 1! + ( n - 1)1!
2019! = 1! + n! - 1!
n! = 2019!
n = 2019
Total sum = 2019/2 ( 1! + 2019!)
= 2019/2 × 2020!
= 2039190!
Now ,
2039190!/21
= 97104whole 6!/21
Hence , 6! is the reminder .
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