Question

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

Answer 1

QUESTION -

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

ANSWER -

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

when r=0

a = 6q,even no

when r=1

a = 6q + 1, odd no

when r=2

a = 6q + 2, even no

when r = 3

a=6q + 3,odd no

when r=4

a=6q + 4,even no

when r=5,

a= 6q + 5 , odd no

...Any positive odd integer is of the form 6q+1,6q+3 or 6q+5

@HarshPratapSingh

Answer 2

$\Large{\underline{\underline{\mathfrak{AnSwEr :}}}}$

Let x be a given integer.

On dividing x by 6 , we get q as the quotient and r as the remainder such that

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

$\rule{100}{2}$

$\sf{When \: r = 0} \\ \\ \sf{\dashrightarrow x = 6q} \\ \\ \bf{Which \: is \: even \: Number}$

$\rule{100}{2}$

$\sf{When \: r = 1} \\ \\ \sf{\dashrightarrow x = 6q + 1} \\ \\ \bf{Which \: is \: odd \:number}$

$\rule{100}{2}$

$\sf{When \: r = 2} \\ \\ \sf{\dashrightarrow x = 6q + 2} \\ \\ \bf{Which \: is \: Even \:number}$

$\rule{100}{2}$

$\sf{When \: r = 3} \\ \\ \sf{\dashrightarrow x = 6q + 3} \\ \\ \bf{Which \: is \: odd \:number}$

$\rule{100}{2}$

$\sf{When \: r = 4} \\ \\ \sf{\dashrightarrow x = 6q + 4} \\ \\ \bf{Which \: is \: Even \:number}$

$\rule{100}{2}$

$\sf{When \: r = 5} \\ \\ \sf{\dashrightarrow x = 6q + 5} \\ \\ \bf{Which \: is \: odd \: number}$

So, Any positive odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.