What are the last two digits of (301*401*503*604*646*547*448*349)^2	Select
A) 96
B) 76
C) 56
D) 44
E) 16
((301*401*503*604*646)*(547*448*349))^2   If you observe above digits, last digit are: 1,1,3,4,6,7,8,9; 1&amp;5 are missing; so I have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01,01,03,04,46 and final three as 47*48*49.   Solving for only last two digits and multiplying them we get:( (03*04*46)(56*49))^2 = (52*44)^2=88^2=44   Hence answer is 44
<result>D</result>
What are the last two digits of (301*401*503*604*646*547*448*349)^2