What are the last two digits of (301*402*503*604*646*547*448*348)^2  Pick one: A) 86 B) 76 C) 56 D) 36 E) 16
((301*402*503*604*646)*(547*448*348))^2   If you observe above digits, last digit are: 1,2,3,4,6,7,8,8; 5 is missing; so I have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01,02,03,04,46 and final three as 47*48*48.   Solving for only last two digits and multiplying them we get:( (06*04*46)(56*48))^2 = (44*88)^2=72^2=84   Hence answer is 86
<result>A</result>
What are the last two digits of (301*402*503*604*646*547*448*348)^2