Use division algorithm to show that the cube of any positive integer is of the form 9 m 9m + 1 or 9m + 8.
Answers
Step-by-step explanation:
Given :-
a positive integer
To find :-
Use division algorithm to show that the cube of any positive integer is of the form 9 m or 9m + 1 or 9m + 8.
Solution :-
Euclid's Division Algorithm:-
For any two positive integers a and b there exist two positive integers q and r satisfying a=bq+r, where 0≤r<b.
Consider a = 9q + r ------------------(1)
Where 0≤r<9
The possible values of r = 0,1,2,3,4,5,6,7,8
Case-1:-
If r = 0 in (1) then
=> a = 9q+0
=> a = 9q
On cubing both sides then
=> a³ = (9q)³
=> a³ = 729q³
=> a³ = 9(81q³)
=> a³ = 9m ------------------(2)
Where ,m = 81q³ ,m is an integer
Case -2:-
If r = 1 in (1) then
=> a = 9q+1
On cubing both sides then
=> a³ = (9q+1)³
=> a³ = (9q)³+3(9q)²(1)+3(9q)(1)²+(1)³
=> a³ = 729q³ + 3(81q²)+27q+1
=> a³ = 729q³ + 243q²+27q+1
=> a³ = 9(81q³+27q²+3q)+1
=> a³ = 9m+1 ------------------(3)
Where ,m = 81q³+27q²+3q ,m is an integer
Case-3:-
If r = 2 in (1) then
=> a = 9q+2
On cubing both sides then
=> a³ = (9q+2)³
=> a³ = (9q)³+3(9q)²(2)+3(9q)(2)²+(2)³
=> a³ = 729q³ + 6(81q²)+27q(4)+8
=> a³ = 729q³ + 486q²+108q+8
=> a³ = 9(81q³+54q²+12q)+8
=> a³ = 9m+8 ------------------(4)
Where ,m = 81q³+54q²+12q ,m is an integer.
Case-4:-
If r = 3 in (1) then
=> a = 9q+3
On cubing both sides then
=> a³ = (9q+3)³
=> a³ = (9q)³+3(9q)²(3)+3(9q)(3)²+(3)³
=> a³ = 729q³ + 9(81q²)+27q(9)+27
=> a³ = 729q³ + 729q²+243q+27
=> a³ = 9(81q³+81q²+27q+3)
=> a³ = 9m------------------(5)
Where ,m = 81q³+81q²+27q+3 ,m is an integer.
Case-5:-
If r = 4 in (1) then
=> a = 9q+4
On cubing both sides then
=> a³ = (9q+4)³
=> a³ = (9q)³+3(9q)²(4)+3(9q)(4)²+(4)³
=> a³ = 729q³ + 12(81q²)+27q(16)+64
=> a³ = 729q³ + 972q²+432q+64
=> a³ = 729q³ + 972q²+432q+63+1
=> a³ = 9(81q³+108q²+48q+7)+1
=> a³ = 9m+1 ------------------(6)
Where ,m = 81q³+108q²+48q+7 ,m is an integer.
Case-6:-
If r = 5 in (1) then
=> a = 9q+5
On cubing both sides then
=> a³ = (9q+5)³
=> a³ = (9q)³+3(9q)²(5)+3(9q)(5)²+(5)³
=> a³ = 729q³ + 15(81q²)+27q(25)+125
=> a³ = 729q³ + 1215q²+675q+125
=> a³ = 729q³ + 1215q²+675q+117+8
=> a³ = 9(81q³+135q²+75q+13)+8
=> a³ = 9m+8 ------------------(7)
Where ,m = 81q³+135q²+75q+13 ,m is an integer.
Case-7:-
If r = 6 in (1) then
=> a = 9q+6
On cubing both sides then
=> a³ = (9q+6)³
=> a³ = (9q)³+3(9q)²(6)+3(9q)(6)²+(6)³
=> a³ = 729q³ + 18(81q²)+27q(36)+216
=> a³ = 729q³ + 1458q²+972q+216
=> a³ = 9(81q³+162q²+108q+24)
=> a³ = 9m ------------------(8)
Where ,m = 81q³+162q²+108q+24 ,m is an integer.
Case-8:-
If r = 7 in (1) then
=> a = 9q+7
On cubing both sides then
=> a³ = (9q+7)³
=> a³ = (9q)³+3(9q)²(7)+3(9q)(7)²+(7)³
=> a³ = 729q³ + 21(81q²)+27q(49)+343
=> a³ = 729q³ + 1701q²+1323q+343
=> a³ = 729q³ + 1701q²+1323q+342+1
=> a³ = 9(81q³+189q²+137q+38)+1
=> a³ = 9m+1 ------------------(9)
Where ,m = 81q³+189q²+137q+38 ,m is an integer.
Case-9:-
If r = 8 in (1) then
=> a = 9q+8
On cubing both sides then
=> a³ = (9q+8)³
=> a³ = (9q)³+3(9q)²(8)+3(9q)(8)²+(8)³
=> a³ = 729q³ + 24(81q²)+27q(64)+512
=> a³ = 729q³ + 1944q²+1728q+512
=> a³ = 729q³ + 1944q²+1728q+504+8
=> a³ = 9(81q³+216q²+192q+56)+8
=> a³ = 9m+8 ------------------(10)
Where ,m = 81q³+216q²+192q+56,m is an integer.
From all above equations from (2) to (10)
We conclude that
The cube of any positive integer is in the from of 9m or 9m+1 or 9m+8
Hence, Proved.
Answer:-
The cube of any positive integer is in the from of 9m or 9m+1 or 9m+8.
Used formulae:-
Euclid's Division Algorithm:-
For any two positive integers a and b there exist two positive integers q and r satisfying a=bq+r, where 0≤r<b.
Used Identity :-
(a+b)³ = a³+3a²b+3ab²+b³