Divisibility Question. Show that if $a$ and $b$ are odd positive integers, and $b$ does not divide $a$, then there are integers $s$ and $t$ such that $ a = bs + t$ where $t$ is odd and $|t| &lt;b$.

Let a = 2k +1 and b = 2j + 1 where k,j are positive integers.

since b does not divide a then there is a non-zero remainder on division by b.
we can write;

$a = bs + t$

now to show that $t$ is odd;

$(2k +1) = (2j + 1)s + t$

rearranging;

$2(k-2js) - ( s-1) = t$

$2(k-2js) + (1-s) = t$

$2[(k-2js) -s/2] + 1  = t$

Hence t is odd. but i have not shown that $|t| &lt;b$.

In fact, i am questioning whether i have shown anything.
The most straightforward method is by using the euclidian algorithm:

So by the euclidian algorithm, we find $q$ and $r$ such that
$$a = bq+r$$
and $0&lt;|r|&lt;|b| = b$. We consider two cases:


$r$ is odd. We are done.
$r$ is even, but in this case, $r-b$ is odd, so $$a = b(q+1)+(r-b)$$ and $-b &lt; r-b &lt; 0$ so $|r-b| &lt; b$.


I think that almost every proof will require a (small) case-study.
Note: The following is more or less a rewrite of JefLaga's answer with a small additional hint in order to answer the detail which is asked by OP in the bounty text.
  
  It's a good idea to start as JefLaga did and use the Euclidean algorithm, because this way we can already assume the existence of non-negative integers $q$ and $r$ such that
  \begin{align*}
a=bq+r\qquad\qquad 0\leq|r|&lt;b
\end{align*}
  Since $b$ does not divide $a$ we observe
  $$0&lt;|r|&lt;b$$
  
  So, the only thing we have to show is that $r$ is odd.


Let's recall the parity when adding/subtracting even and odd numbers:
\begin{array}{lcll}
\pm&amp;|&amp;even&amp;odd\tag{1}\\
\hline&amp;|\\
even&amp;|&amp;even&amp;odd\\
odd&amp;|&amp;odd&amp;even
\end{array}


  And as Jef did, we distinguish according to the parity of $r$
  
  
  $r$ is odd: Nothing to show
  $r$ is even: Now since $b$ is odd, we observe according to (1) that $r-b$ is odd (even minus odd = odd)
  
  
  and again we have a valid representation
  $$a=b(q+1)+(r-b)$$
  with an odd number $r-b$ as remainder and since $-b&lt;r-b&lt;0$ we observe: 
  \begin{align*}
|r-b|&lt;b\qquad\qquad&amp;\\
&amp;\Box
\end{align*}