 
<translate> Nchịkọta # nke koodu nwere ike ịbụ 3*3*3 = 27. N'ime koodu iri abụọ na asaa ahụ, naanị mkpụrụ nke 333 ga-abụ nke na-adịghị ahụkebe, 26 ndị fọdụrụ ga-enwe 2 ma ọ bụ 4 n'ime ha, nke na-eme ka o doo anya na mkpụrụ ha ga-abụ ọbụna. N'ihi ya, ọnụ ọgụgụ nke koodu ebe ngwaahịa nke digits bụ ọbụna = (ngụkọta) - (mgbochi) = 27 - 1 = 26. Azịza ya bụ: (B). <eng> Total # of codes possible is 3*3*3 = 27. Oit of those 27 codes only the product of 333 will be odd, the remaining 26 will have either 2 or 4 in them, which ensures that their product will be even. Therefore the number of codes where the product of the digits is even = (total) - (restriction) = 27 - 1 = 26.
The answer: (B). 