Question

Use euclid division lemma ro show that the cube of any positive integer is of the form 9m, 9m+1,9m+8

Answer 1

let any positive integer be a .
by euclid division lemma ,
a= bq+ r

let it be divided by 3 .
so , a = 3q+ r
now possible remainder = 0,1,2

Case 1 ) when remainder is 0 ,
a = 3q+0 => a= 3q
cubing both sides ,
a³ = (3q)³
=> a³ = 27q³
=> a³ = 9(3q³)
let 3q³ be m .

so a = 9m
Case 2) when remainder is 1
similarly , a = 3q+1
=> a³ = (3q+1)³ [ cubing both sides ]
=> a³= 9q³+1+18q²+9q
=> a³= 9(q³+2q²+q) + 1
let (q³+2q²+q) be m
so , a³= 9m+1 .

case 3) .when remainder is 2
a = 3q+2 .
a³ = 9q³+18q²+9q+8 [ cubing bith sides ]
a³= 9(q³+2q²+q) +8
ket (q³+2q²+q) be m
so , a³= 9m+8 .

so we can say by there three cases that the cube if any positive integer is in the firm of ( 9m ), (9m+1) , ( 9m+8)

Answer 2

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 .