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Show that any positive odd integer is of the form $$6q+1,6q+3$$ or $$6q+5,$$ where $$q$$ is some integer.
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## QuestionMathsClass 10

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

See solution below
4.6
4.6

## Solution

Let $$a$$ be any positive odd integer.
On taking, $$b=6$$ and on applying Euclid's division lemma on $$a$$ and $$6$$ there exist unique integers $$q$$ and $$r$$ such that:
$$a=6q+r\;\;\;[0\leq r< 6\;\mathrm{i.e,}\;r =0 ,1,2,3,4,5]$$
If $$r=0,\Rightarrow a=6q,$$ and  $$6q = 2(3q)$$ is divisible by 2
$$\Rightarrow$$ $$6q$$is even.
If $$r=1,\Rightarrow a=6q+1$$ and $$6q+1$$ is not divisible by 2.
If $$r=2,\Rightarrow a=6q+2$$ and $$6q+2=2(3q+1)$$ is divisible by 2
$$\Rightarrow 6q+2$$ is even.
If $$r=3,\Rightarrow a=6q+3,$$ and $$6q+3$$ is not divisible by 2.
If $$r=4,\Rightarrow a=6q+4,$$ and $$6q+4=2(3q+2)$$ is divisible by 2
$$\Rightarrow$$ $$6q+4$$ is even.
If $$r=5,\Rightarrow a=6q+5$$ and $$6q+5$$ is not divisible by 2.
Since, $$6q,6q+2,6q+4$$ are even . So these cases will not be considered as a is an odd positive integer and the remaining cases $$6q+1,6q+3$$ and $$6q+5$$are odd.
So, we can conclude that all the positive odd integers will be of the form of $$6q+1,6q+3$$ or $$6q+5$$