Question

Show that any positive odd integer is of the form 6q+1, or 69 +3, or 69 + 5, where q
some integer.
​

Answer 1

lets us start with taking a,where a is a postive off integer

We apply the division algorithm with a and b=6

since 0 less than or equal to r or lessthan 6 the possible remainders are 012345

that is

a can be

6q+1or6q+3or6q+5

where quotient

however since a is odd a cannot be 6q,6q+2,6q+4since they are divisble by 2

therefore the any odd integer is of the form of 6q+1,6q+3,6q+5

Thank You

Answer 2

$\underline\mathfrak\purple{By\:Euclid\:Division\:Lemma\::-}$

if 'a' and 'b' are two positive integers then

$\fcolorbox{red}{yellow}{a\:=\:bq+r}$

where,

0 ≤ r < b

let a be a positive integers and b = 6

$\fcolorbox{red}{yellow}{a\:=\:6q+r}$

where,

0 ≤ r < 6

from above we can say that remainder is less than 6 and equal or greater than 0

so the possible values of r be

$\mathfrak\green{0,1,2,3,4,5}$

by putting values of divisor(b) and remainder in equation $\fbox\color{red}{a\:=\:bq+r}$

putting r = 0, a = 6q + 0

a = 6q

it is divided by 2 so it is an $\underline\mathfrak\blue{even}$ number.

putting r = 1, a = 6q + 1

6q is divided by 2 but 1 is not divided.

it is not divided by 2 so it is an $\underline\mathfrak\blue{odd}$ number.

putting r = 2, a = 6q + 2

it is divided by 2 so it is an $\underline\mathfrak\blue{even}$ number.

putting r = 3, a = 6q + 3

it is not divided by 2 so it is an $\underline\mathfrak\blue{odd}$ number.

putting r = 4, a = 6q + 4

it is divided by 2 so it is an $\underline\mathfrak\blue{even}$ number

putting r = 5, a = 6q + 5

it is not divided by 2 so it is an $\underline\mathfrak\blue{odd}$ number.

so by these we can say that any positive odd integer is of the form 6q+1, or 69 +3, or 69 + 5, where q is some integer.

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