Question

[3^3x3+2^45:2^42+(3^5)^6:3^29+9^0]:31

Answer 1

[3^3x3+2^45:2^42+(3^5)^6:3^29+9^0]:31 = 
[3^(3+1)+2^(45-42)+3^(6*5):3^29+1]:31=
(3^4+2^3+3^30:3^29+1):31 = 
(3^4+2^3+3^1+1):31 = 
(81+8+3+1):31 = 
91:31 = 3