*
* Suppose %$x = x' + z 2^k$%, where %$k \ge n$%; then
* %$x \equiv x' + d z 2^{k-n} \pmod p$%. We can use this to trim the
- * representation of %$x$%; each time, we reduce %$x$% by a mutliple of
+ * representation of %$x$%; each time, we reduce %$x$% by a multiple of
* %$2^{k-n} p$%. We can do this in two passes: firstly by taking whole
* words off the top, and then (if necessary) by trimming the top word.
* Finally, if %$p \le x < 2^n$% then %$0 \le x - p < p$% and we're done.
* a perfect power of two, and %$d = 0$%, so again there is nothing to do.
*
* In the remaining case, we have decomposed @x@ as %$2^{n-1} + d$%, for
- * some positive %$d%, which is unfortuante: if we're asked to reduce
+ * some positive %$d%, which is unfortunate: if we're asked to reduce
* %$2^n$%, say, we'll end up with %$-d$% (or would do, if we weren't
* sticking to unsigned arithmetic for good performance). So instead, we
* rewrite this as %$2^n - 2^{n-1} + d$% and everything will be good.