Use Euclids division lemma to show that the cube of any positive integer is the form of 9m,9m+1 or 9m+8
Answers
Step-by-step explanation:
How do you show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8?
Invest INR 18000 per month and get INR 2 crores on maturity.
Dean Rubine
Answered 1 year agoLuís Sequeira
How do you show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8?
There’s an important fact about integers that makes modular arithmetic work. Here we’re interested in (mod9) , essentially remainders when we divide by 9. For the arithmetic operations addition, subtraction and multiplication, if we’re only interested in the remainder, we can just replace each number by a small number with the same remainder without changing the result. Let’s look at some examples,
82+100≡182≡2(mod9)
82+100≡1+1≡2(mod9)
112≡121≡4(mod9)
112≡22≡4(mod9)
What’s really going on is we’re working in the ring Z9 where the only numbers are 0,1,2,3,4,5,6,7,8.
That’s all to say if we want to verify some property about the remainders of “all numbers” when we divide by some number, we really only have a finit
How does one prove that the cube of any integer has one of the forms 9a, 9a+1 or 9a+8?
1,862 Views
How do you use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1, or 9M+8?
786 Views
How can I use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m + 1? I wanted to know how this can be done and why b = 3.
2,993 Views
How do you use the division algorithm to show that the cube of any positive integer is of the form 3p, 3p+1,3p+8?
325 Views
What is the cube of 1?
17,289 Views
Ajesh K C
Answered 1 year ago
How do you show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8?
Before going to problem lets look in to a concept
Euclid’s Division Lemma
According to this if and b are two positive integers then there exist unique integers q and r such that
a = bq+r where 0 ≤ r < b
For example
Let a = 7 and b = 4. a/b gives a remainder 3
So
a = bq+r
7 = 4*1 + 3
Now consider the problem
x be a real number divided by 3. so x can be represented in three cases according to Euclid’s Division Lemma (as per condition 0 ≤ r < b where b =3 and r takes values , 0, 1 and 2
x = 3q
x = 3q+1
x= 3q +2
CASE I
x = 3q
x^3 = (3q)^3 =27q^3 = 9*(3*q^3)
Let m = 3*q^3
x^3 =9m
CASE II
x = 3q+1
x^3 = (3q+1)^3 = (3q)^3 +3*(3q)^2*1+3*3q*1^2 + 1^3
= 27q^3 +27q^2 +9q+1
= 9(3q^3+3q^2+q)+1
x^3 = 9m +1
where m = (3q^3+3q^2+q)
CASE III
x = 3q+2
x^3 = (3q+2)^3
= 27q^3 +54q^2 +36q +8
= 9(3q^3+6q^2+4q) + 8
= 9m +8
where m = 3q^3+6q^2+4q)
x^3 = 9m + 8
So cube of any positive integer can be represented in the form 9m, 9m + 1 or 9m + 8
Please mark me as Brainliest
Thank you