Use division algorithm to show that the cube of any positive integer is of the form 9m,9m+1,9m+8.
Answers
Use ED lemma has
a=bq+r ,a=3q+r,
a³=3q³a=27q³,9(3q³)
,9m where m=3q³
Is r=1,a=3q+1
a³=(3q+1³)=(3q)³+(3q)²(1)+3(3q)(1)²+1³
=27q³+9q²+9q
=9(3q³+q²+q) it gives 9m+1
Then a³=(3q+2)
=(3q³)+3q(3q²)(2)+3(3q)(2)²
=27q³+54q²+36q+8
=9(3q³+6q²+4q)+8
=9m+8 where m = (3q³+6q+4q)
from the above we find that the cube of any positive integer is of the form 9m, 9m+1,9m+8
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 .