☆show that cube of every positive integer is in the form of 6p, 6 p + 1 6p - 1 60 - 2 ,6 p + 2 ,6p + 3
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Read the question carefully there is no even or odd Integers mentioned!
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Answers
Step-by-step explanation:
Using Euclid division algorithm, we know that a=bq+r, 0≤r≤b ----(1)
Let a be any positive integer and b=6.
Then, by Euclid’s algorithm, a=6q+r for some integer q≥0, and r=0,1,2,3,4,5 ,or 0≤r<6.
Therefore, a=6qor6q+1or6q+2or6q+3or6q+4or6q+5
6q+0:6 is divisible by 2, so it is an even number.
6q+1:6 is divisible by 2, but 1 is not divisible by 2 so it is an odd number.
6q+2:6 is divisible by 2, and 2 is divisible by 2 so it is an even number.
6q+3:6 is divisible by 2, but 3 is not divisible by 2 so it is an odd number.
6q+4:6 is divisible by 2, and 4 is divisible by 2 so it is an even number.
6q+5:6 is divisible by 2, but 5 is not divisible by 2 so it is an odd number.
And therefore, any odd integer can be expressed in the form 6q+1or6q+3or6q+5
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Step-by-step explanation:
Given :-
A positive integer
To find :-
Show that cube of every positive integer is in the form of 6p, 6 p + 1, 6p - 1 ,6p - 2 ,6 p + 2 ,6p + 3?
Solution:-
We know that
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.
Let b = 6 then a = 6q+r -----------(1)
Where , 0≤r<6
The possible values of r = 0,1,2,3,4,5
i) If r = 0 then (1) becomes
=> a = 6q+0
=> a = 6q
On cubing both sides then
=> a³ = (6q)³
=>a³ = 216q³
=> a³ = 6(36q³)
=> a³ = 6p ------------------------------(2)
where p = 36q³
ii) If r = 1 then (1) becomes
=> a = 6q+1
On cubing both sides then
=> a³ = (6q+1)³
=> a³ = 216q³+3(36q²)(1)+3(6q)(1²)+1
=> a³ = 216q³+108q²+18q+1
=> a³ = 6(36q³+18q²+3q)+1
=> a³ = 6p+1 ---------------------------(3)
Where p = 36q³+18q²+3q
iii) If r = 2 then (1) becomes
=> a = 6q+2
On cubing both sides then
=> a³ = (6q+2)³
=> a³ = 216q³+3(36q²)(2)+3(6q)(2²)+2³
=> a³ = 216q³+216q²+72q+8
=> a³ = 216q³+216q²+72q+6+2
=> a³ = 6(36q³+36q²+12q+1)+2
=> a³ = 6p+2 ----------------------------(4)
Where p = 36q³+36q²+12q+1
iv)If r = 3 then (1) becomes
=> a = 6q+3
On cubing both sides then
=> a³ = (6q+3)³
=> a³ = 216q³+3(36q²)(3)+3(6q)(3²)+3³
=> a³ = 216q³+324q²+162q+27
=> a³ = 216q³+324q²+162q+24+3
=> a³ = 6(36q³+54q²+27q+4)+3
=> a³ = 6p+3 ---------------------------(5)
Where p = 36q³+54q²+27q+4
v)If r = 4 then (1) becomes
=> a = 6q+4
On cubing both sides then
=> a³ = (6q+4)³
=> a³ = 216q³+3(36q²)(4)+3(6q)(4²)+4³
=> a³ = 216q³+432q²+204q+64
=> a³ = 216q³+432q²+204q+66-2
=> a³ = 6(36q³+72q²+34q+11)-2
=> a³ = 6p-2 -----------------------------(6)
Where p = 36q³+72q²+34q+11
vi)If r = 5 then (1) becomes
=> a = 6q+5
On cubing both sides then
=> a³ = (6q+5)³
=> a³ = 216q³+3(36q²)(5)+3(6q)(5²)+5³
=> a³ = 216q³+540q²+750q+125
=> a³ = 216q³+540q²+750q+126-1
=> a³ = 6(36q³+90q²+125q+21)-1
=> a³ = 6p-1 ------------------------------(7)
Where p = 36q³+90q²+125q+21
From (2),(3),(4),(5),(6),(7)
The cube of every positive integer is in the form of 6p,6p+1,6p-1,6p-2,6p+2,6p+3
Hence, Proved.
Answer:-
The cube of every positive integer is in the form of 6p, 6 p + 1, 6p - 1 ,6p - 2 ,6 p + 2 ,6p + 3.
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".