Question

show that any positive odd integer is of the form of 6q+1, 6q+3, 6q+5, where q is some integer

Answer 1

Answer:

Let a be a given integer.

Let a be a given integer.On dividing a by 6 , we get q as the quotient and r as the remainder such that

=> a = 6q + r, r = 0,1,2,3,4,5

a = 6q + r, r = 0,1,2,3,4,5when r=0

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even no

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd no

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even no

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3a=6q + 3,odd no

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3a=6q + 3,odd nowhen r=4

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3a=6q + 3,odd nowhen r=4a=6q + 4,even no

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3a=6q + 3,odd nowhen r=4a=6q + 4,even nowhen r=5,

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3a=6q + 3,odd nowhen r=4a=6q + 4,even nowhen r=5,a= 6q + 5 , odd no

a = 6q + r, r = 0,1,2,3,4,5when r=0a = 6q,even nowhen r=1a = 6q + 1, odd nowhen r=2a = 6q + 2, even nowhen r = 3a=6q + 3,odd nowhen r=4a=6q + 4,even nowhen r=5,a= 6q + 5 , odd noAny positive odd integer is of the form 6q+1,6q+3 or 6q+5.

Answer 2

Answer:

Let 'a' be positive integers ;by Eculid lemma

a=6q,6q+1,6q+2,6q+3,6q+4,6q+5

a=6q=2×(3q) = divisible by 2 =even

a=6q+1 =2×(3q)+1 =not divisible by 2=odd

a=6q+2=2×(3q+1) =divisible by 2=even

a=6q+3=2×(3q) +3=not divisible by 2=odd

a=6q+4=2×(3q+2)= divisible by 2=even

a=6q+5=2×(3q)+5 =not divisible by 2=odd

so, the odd forms a

6q+1,6q+2,6q+5 are positive integers