SOLUTION: how will I show that the cube of any number is in the form 7k or 7k+1 or 7k-1
Algebra.Com
Question 462731: how will I show that the cube of any number is in the form 7k or 7k+1 or 7k-1
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
Every integer falls into one of the following modulo classes
7n
7n + 1
7n + 2
7n + 3
7n + 4
7n + 5
7n + 6
For the first class, , which simply says that the cube of any multiple of 7 is still divisible by 7.
For the 2nd class 7n + 1,
==> .
For the 3rd class 7n + 2,
==> .
For the 4nd class 7n + 3,
==> .
For the 5th class 7n + 4,
==> .
For the 6th class 7n + 5,
==> .
For the 7th class 7n + 6,
==> .
In other words, the cube of any integer is either of the form 7k, 7k +1, or 7k + 6. But numbers of the form 7k + 6 are the same numbers of the form 7k - 1. The proof is complete.
RELATED QUESTIONS
verify that if an integer is simultaneously a square and a cube , then it must be either... (answered by richard1234)
verify that if an integer is simultaneously a square and a cube then it must beveither of (answered by richard1234,Edwin McCravy)
3k-1=7k+2 (answered by jim_thompson5910)
3k-1=7k+2 (answered by Boreal)
What is -1 - 8(1 + 7k) simplified? (answered by lynnlo)
7k^2-5k-1=0 (answered by TX911)
5k+34=-2(1-7k) (answered by tommyt3rd)
Find the value of k so that 2k + 2, 5k - 11, and 7k - 13 will form an arithmetic... (answered by srinivas.g,MathTherapy)
.Find k so that the numbers 2k+1, 3k+4, and 7k+6 form a geometric
sequence
(answered by jim_thompson5910)