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

Answers (1)

According to Eudid Division Lomma

if a,b are positive integers, then,

$a=bq+r,$ where $0\leq r< b$

Assume $a=$ positive integer and $b=6$

$\Rightarrow a=6q+r$

and here $0\leq r\; <6\Rightarrow \left \{ r=0,1,2,3,4,5 \right \}$

if $r=0\Rightarrow a=6q\Rightarrow$ it is even

if $r=1\Rightarrow a=6q+1\Rightarrow$ it is odd

if $r=2\Rightarrow a=6q+2\Rightarrow$ it is even

if $r=3\Rightarrow a=6q+3\Rightarrow$ it is odd

if $r=4\Rightarrow a=6q+4\Rightarrow$ it is even

if $r=5\Rightarrow a=6q+5\Rightarrow$ it is odd

Hence, we can infer that the number will in the form of

$6m+1,6m+3, \; or \; 6m+5$

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Safeer PP

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