What are the last two digits of (301*402*502*604*646*547*448*349)^2
What are the last two digits of (301*402*502*604*646*547*448*349)^2
Answer/Solution
56
Steps/Work
((301*402*502*604*646)*(547*448*349))^2
If you observe above digits, last digit are: 1,2,2,4,6,7,8,9; 3&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 01,02,02,04,46 and final three as 47*48*49.
Solving for only last two digits and multiplying them we get:( (04*04*46)(56*49))^2 = (36*44)^2=84^2=56
Hence answer is C
If you observe above digits, last digit are: 1,2,2,4,6,7,8,9; 3&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 01,02,02,04,46 and final three as 47*48*49.
Solving for only last two digits and multiplying them we get:( (04*04*46)(56*49))^2 = (36*44)^2=84^2=56
Hence answer is C