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 $|t| <b$.
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 $|t| <b$.
In fact, i am questioning whether i have shown anything. 


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|<b
\end{align*}
  Since $b$ does not divide $a$ we observe
  $$0<|r|<b$$
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&|&even&odd\tag{1}\\
\hline&|\\
even&|&even&odd\\
odd&|&odd&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 $-b<r-b<0$ we observe: 
  \begin{align*}
|r-b|<b\qquad\qquad&\\
&\Box
\end{align*}

