find unit digit of (1!)^100!+(2!)^99!+(3!)^98!+-------(100!)^1!
Answers
9 would be unit digit of (1!)^100!+(2!)^99!+(3!)^98!+-------(100!)^1!
Step-by-step explanation:
(1!)^100!+(2!)^99!+(3!)^98!+-------(100!)^1!
(1!)^100! will end with 1
(2!)^99! -
99! has 4 as one factor
2! = 2
2⁴ = 16 any number ending with 6 raised to power any number ≥ 1 will end with 6
(2!)^99! will end with 6
(3!)^98!
= 6^98!
Hence (3!)^98! will end with 6
(4!)^97!
= 24^97!
97! has 2 as one factor
24² will end with 6
Hence (4!)^97! will end with 6
now 5! = 120 end with 0 & all further number will ends with 0 as all has 5! as one factor
so all further digits ends with 0
Hence
unit digit would be
1 + 6 + 6 + 6 + 0
= 19
9 would be unit digit of (1!)^100!+(2!)^99!+(3!)^98!+-------(100!)^1!
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