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When 1!+2!+3!+4!+5!....(95)! is divided by 15all the factorails from 5! + 6! + 7! + ... (95)! are divisible by 15 as they contain 3*5 as one of their factor.
Thus the remainder can be found by dividing 1!+2!+3!+4! by 15 = 33 by 15. = 3
Regards
Arun
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