What are the last two digits of (301*402*503*604*645*547*448*349)^2	Answers
A) 96 B) 76 C) 56 D) 36 E) 00
((301*402*503*604*645)*(547*448*349))^2   If you observe above digits, last digit are: 1,2,3,4,5,7,8,9; 6 is missing; so I have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01,02,03,04,45 and final three as 47*48*49.   Solving for only last two digits and multiplying them we get:( (06*04*45)(56*49))^2 = (00*44)^2=00^2=00   Hence answer is 00
<result>E</result>
What are the last two digits of (301*402*503*604*645*547*448*349)^2