use division algorithm to show that the cube of any positive integer is of the form 9m or 9m+1 or 9m+8
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Answered by
18
We know,
For every positive integer a and b there always exists another two integers q and r such that a=bq+r and 0<=r<b.
Taking b=3
a=3q+0 or
a=3q+1 or
a=3q+2
Now,
a^3=(3q)^3
=27q^3
=9×3q^3
=9m[Taking 3q^3=m]
a^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
=9m+1[Taking 3q^3+3q^2+q=m]
a^3=(3q+2)^3
=(3q)^3 + 3×(3q)^2×2 + 3×3q×2^2+2^3
=27q^3 + 54q^2 +36q+8
=9(3q^3+6q^2+4q)+8
=9m+8[Taking 3q^3+6q^2+4q=m]
proved....
hope it helps u....
For every positive integer a and b there always exists another two integers q and r such that a=bq+r and 0<=r<b.
Taking b=3
a=3q+0 or
a=3q+1 or
a=3q+2
Now,
a^3=(3q)^3
=27q^3
=9×3q^3
=9m[Taking 3q^3=m]
a^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
=9m+1[Taking 3q^3+3q^2+q=m]
a^3=(3q+2)^3
=(3q)^3 + 3×(3q)^2×2 + 3×3q×2^2+2^3
=27q^3 + 54q^2 +36q+8
=9(3q^3+6q^2+4q)+8
=9m+8[Taking 3q^3+6q^2+4q=m]
proved....
hope it helps u....
Answered by
10
Step-by-step explanation:
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ r = 0,1,2 .
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
Where m is an integer such that m =
Case 2: When a = 3q + 1,
a = (3q +1) ³
a = 27q ³+ 27q ² + 9q + 1
a = 9(3q ³ + 3q ² + q) + 1
a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .
Case 3: When a = 3q + 2,
a = (3q +2) ³
a = 27q³ + 54q² + 36q + 8
a = 9(3q³ + 6q² + 4q) + 8
a = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Hence, it is proved .
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