Pramod J AskiitiansExpert-IIT-B
Last Activity: 13 Years ago
Dear student,
any number is either 2n or 2n+1.
2n^2 = 4n^2 so this is divisible by 4
(2n+1)^2 = 4n^2 + 4n + 1
in the above term 4n^2 and 4n are divisible by 4 and leaves a reminder of 1
similarly for a cube of an interger(which can be expressed in one of these forms 3n or 3n+1 or 3n+2) divided by 9
[3n]^3 = 27n^3 when divided by 9 leaves 0
[3n+1]^3 = 27n^3 + 27n + 9n^2 + 1 when divided by 9 leaves 1
[3n+2]^2 = 27n^3 + 36n + 18n^2 + 8 when divided by 9 leaves 8
i hope this explains !!