Numbers : Remainder 102

(1!)³+(2!)³+(3!)³+.............+(1152!)³/(1152!)        remainder ?


Q. (1!)³+(2!)³+(3!)³+.............+(1152!)³/(1152)

Ans:

Divisor = 1152 = 2⁷ * 3²

(1!)³ = 1 % 1152 = 1

(2!)³ = 8 % 1152 = 8

(3!)³  = 216 % 1152 = 216

(4!)³ = (24)³ = (2⁹  * 3³ ) % 1152 = 0 

By observation, for higher factorial value you will find that remainder with 1152 is 0.................(since power of 2 and 3 will be more than 7 and 2 respectively) 

OR

You can say, 4! will occur @ higher factorial value which is already divisible by 1152 hence the remainder with 1152 will be 0. 

Hence the remainder = 1 + 8 + 216 = 225