July 2002
Intermediate to advanced
320 pages
8h 15m
English
At this point you may wonder if other divisors present other problems. We see in this section that they do not; the three examples given illustrate the only cases that arise (for d ≥ 2).
Some of the proofs are a bit complicated, so to be cautious, the work is done in terms of a general word size W.
Given a word size W ≥ 3 and a divisor d, 2 ≤ d < 2W − 1, we wish to find the least integer m and integer p such that
Equation 1a
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Equation 1b
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with 0 ≤ m < 2W and p ≥ W.
The reason we want the least integer m is ...
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