Digit at the unit place of the sum of 1 ! 2 + 2 ! 2 + 3 ! 2 … … … + 2008 ! 2 is

Step-by-step solution

Given the sum S = ( 1 ! ) 2 + ( 2 ! ) 2 + ( 3 ! ) 2 + ....... + ( 2008 ! ) 2 Since the digit at units place is zero in n ! for n ≥ 5 Hence, S = ( 1 ) 2 + ( 2 ) 2 + ( 6 ) 2 + ( 24 ) 2 + numbers having zero at units place = 617 + all other numbers having zero at units place