The remainder when 1! + 2! + 3!...+ 99! divided by 10???
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After 3! all the number is divisible by 12 .
From 4! onwards all the terms have the terms with makes 4! Perfectly divisible by 12 .
This makes 4!+……..+100! Term remainder is 0 .
Now for 1! to 3! They have insufficient terms to be a multiple of 12 and also less than 12 . So 1!+2!+3!=9 will remain only , hence the answer will be 9.
Step-by-step explanation:
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