The unit digit of expression (1!)^99 + (2!)^98 + (3!)^97+ .........(99!)^1 is(A)1 (b)3 (c)7 (d)6

The unit digit of (1!)^99 is 1.The unit digit of (2!)^98 is 4.The unit digit of (3!)^97 is 6.The unit digit of (3!)^97 is 6.The unit digit of (4!)^96 is 0.Now since we know that all the factorial numbers starting with 5! has its unit digit 0.So we need not to calculate it.Thus the required unit digit= unit digit of the sum of the unit digits.Thus the unit digit is 7   (1+4+6+6+0=17). Hope you got your answer ☺😊