Question

Use Euclid's division Lemma to show that the cube of any positive integer is either of the form 9m,9m+1,9m+8 for integer m

Answer 1

for given positive integers a and b there exists unique integers q and r satisfing a=bq+r,0<b≤r.
a=bq+r. —1
put b=9in 1
hence, a=9q+r where r=0,1,2,3,4,5,6,7or8
if r=0,a=9q
a³=(9q) ³=729q³=9(81q³)=9m,
where m=81q³
if r=1,a=9q+1
a³=(9q+1)³
=729q³+243q²+27q+1
=9(81q³+27q²+3q) +1
=9m+1,where m=81q³+27q²+3q
if r=2,a=9q+2
a³=(9q+2)³
=729q³+486q²+108q+8
=9(81q³+56q²+12q)+8
=9m+8,where m=81q³+56q²+12q
similarly,
if r=3,a³=9m
if r=4,a³=9m+1
if r=5,a³=9m+8
if r=6,a³=9m
if r=7,a³=9m+1
if r=8,a³=9m+8
therefore, the cube of any positive integer is either of the form 9m, 9m+1,9m+2 for some integer m.

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 .

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