use Euclid's division lemma to show that the cube. of any positive integer is of the form 9m,9m+1 or 9m+8
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According to Euclid's Division Lemma,
a = bq + r,
Let 'a' be any positive integer , then it is of the form 3q , 3q + 1, 3q +2
Here are the following cases ----
CASE I ---> When a = 3q
=>a = 3q
=> a^3 = (3q)^3
=> a^3 = 27q^3
=> a^3 = 9(3q^3) = 9m, where m = 3q^3
CASE II ---> When a = 3q + 1
=> a = 3q + 1
=> a^3 = (3q + 1)^3
=> a^3 = 27q^3 + 27q^2 + 9q + 1
=> a^3 = 9m + 1 ,
where m = q(3q^2 + 3q + 1)
CASE III ---> When a = 3q + 2
=> a = 3q + 2
=> a^3 = (3q + 2)^3
=> a^3 = 27q^3 + 54q^2 + 36q + 8
=> a^3 = 9m + 8 ,
where m = q(3q^2 + 6q + 4)
Therefore, a^3 is either of the form 9m, 9m + 1 or 9m + 8....
a = bq + r,
Let 'a' be any positive integer , then it is of the form 3q , 3q + 1, 3q +2
Here are the following cases ----
CASE I ---> When a = 3q
=>a = 3q
=> a^3 = (3q)^3
=> a^3 = 27q^3
=> a^3 = 9(3q^3) = 9m, where m = 3q^3
CASE II ---> When a = 3q + 1
=> a = 3q + 1
=> a^3 = (3q + 1)^3
=> a^3 = 27q^3 + 27q^2 + 9q + 1
=> a^3 = 9m + 1 ,
where m = q(3q^2 + 3q + 1)
CASE III ---> When a = 3q + 2
=> a = 3q + 2
=> a^3 = (3q + 2)^3
=> a^3 = 27q^3 + 54q^2 + 36q + 8
=> a^3 = 9m + 8 ,
where m = q(3q^2 + 6q + 4)
Therefore, a^3 is either of the form 9m, 9m + 1 or 9m + 8....
mysticd:
=9m+1
am i right?
plz , edit 2 lines
1) in case II , last line
where m =9(9q^3+3q^2+q)
and
2) in cas III, last line
whare m = 3q^3+6q2+4q
no problem , It's ok.
Answered by
2
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
\ a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms.
We have three cases.
Case 1: When a = 3q,
Where m is an integer such that m =3q3
Case 2: When a = 3q + 1,
Where m is an integer such that
Case 3: When a = 3q + 2,
Where m is an integer such that
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
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