Help with this number theory problem!!! problem!?

Question

Use the division algorithm to establish that the cube of any integer has one of the forms 9k, 9k+1, 9K+8

Answer 1

Every integer is in exactly one of the forms:

n = 3*m
n = 3*m+1
n = 3*m+2

Cubing each expression, we get:
n^3 = (3*m)^3 = 27*m^3 = 9*(3*m^3) = 9*k, k = 3*m^3

n^3 = (3*m+1)^3 = 27*m^3+27*m^2+9*m+1 =
= 9*m*(3*m^2+3*m+1)+1 = 9*k+1, k = m*(3*m^2+3*m+1)

n^3 = (3*m+2)^3 = 27*m^3+18*m^2+36*m+8 =
= 9*m*(3*m^2+2*m+4)+8 = 9*k+8, k = m*(3*m^2+2*m+4)

Don't know what division algortihm is, but this is a proof!

Answer 2

The Division Algorithm states that given any integer n, and any positive integer dividor b of n, we can write

n = bq + r,

where q is an integer (called the "quotient"), and r is an integer greater than or equal to 0 but less than b (called the "remainder").

We will apply this algorithm with b = 3, so that the remainder must be 0, 1, 0r 2. Thus, every integer n has the form

n=3q + r,

where r is one of the numbers 0, 1, or 2. It follows from the Binomial Theorme that

n^3 = 27q^3 + 27(q^2)r + (9q)r^2 + r^3

=9(3q^3 + 3rq^2 + qr^2) + r^3,

where r = 0, 1, or 2.

Therefore, the result is true with

k = 3q^3 + 3rq^2 + qr^2 ,

and with r^3 being one of 0, 1, or 8.

Answer 3

any int can be written as 3k+r,where r=0,1, 2...right
then the cube ;
(3k+r)^3=3k^3+3(3k^2)*r+3(3k*(r^2))+r^3
=27*k^3+27*k^2*r+9k*r^2+r^3
from the first 3 terms u can take a 9 out;
then u'll get something like...9X+r^3
since we toke r to be 0,1,2
each of the cubes of r well give 0,1,8

Answer 4

Even Jesus hates you.