0 votes 0 votes What is −6 mod 18? Please elaborate. Mizuki asked Nov 20, 2018 Mizuki 274 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply ankitgupta.1729 commented Nov 20, 2018 reply Follow Share It is $((-6) + 18)\; mod \; 18 =12$ $x \;mod\; n = (x + n)\; mod \;n = (x + 2n)\; mod n = ....= (x+kn)\; mod\; n$ , where $k$ is an integer(either positive , negative or zero) We can verify it by $(a+b)\; mod\; n \;=(a \;mod \;n + b\; mod \;n)\; mod\; n$. $mod$ $n$ represents total $n$ equivalence classes(partitions) and each class represents remainders $0,1,2,....,(n-1)$ Suppose , we have to do $mod \;3$ operation on numbers then possible remainders will be $0,1,2$ {$0,1,2$} Now add 3 to each number in the set. {$3,4,5$} Now again add 3 to each number in the set. {$6,7,8$} Now again add 3 to each number in the set. {$9,10,11$} .................... {$3k , 3k+1,3k+2$}, where , $k$ is an integer Here , we have 3 equivalence classes. 1) {$.....-6,-3,0,3,6,9,...,3k,...$} represents remainder $0$ class 2) {$.....-5,-2,1,4,7,10,...,3k+1,...$} represents remainder $1$ class 3) {$.....-4,-1,2,5,8,11,...,3k+2,...$} represents remainder $2$ class In each set , every number comes after a fixed interval of $3$. Now, here, for mod 18 , possible remainders are from 0 to 17... Suppose , we have to find the set of numbers which gives remainder 12. So , we have to add/subtract a fixed interval 18 to each number in the set here.Since , 12 will definitely will be in the set. So, set will be {$...,-24,-6,12,30,48,....$} So, $-6 \;mod \;18$ is same as $12\; mod\; 18$ which is also same as $30\; mod \;18$ because all are giving same remainder. 2 votes 2 votes Mizuki commented Nov 20, 2018 reply Follow Share Thank you very much @ankitgupta.1729 for explaining! 1 votes 1 votes Please log in or register to add a comment.