Question

The remainder when 1! + 2! + 3!...+ 99! divided by 10???​

Answer 1

Answer:

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: