* in the correct multiple to subtract, we can shift a byte at a time.
* This produces a 40-bit (rather than a 33-bit) intermediate remainder,
* but again the multiple of the polynomial to subtract depends only on
- * the high bits, the high 8 bits in this case.
+ * the high bits, the high 8 bits in this case.
*
* The multiple we need in that case is the low 32 bits of a 40-bit
* value whose high 8 bits are given, and which is a multiple of the