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RE: solutions

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Dave Hoover
January 28, 2002 20:08
RE: solutions
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Rick Myers wrote:
>On Jan 28, 2002 at 19:56:05 -0600, Dave Hoover wrote:
>> Can someone explain to my poor ignorant head what "low-order bit" means?
>> I'm guessing it has something to do with the '&' operator.  Be gentle.
>Simple. As each requirement for this hole was supposed to be
>"even", it had to be a multiple of two. Thus, in binary
>terms, the "low-order-bit" would always be zero.
>Of course though, if you don't grok binary, that probably
>doesn't make any sense.

push @things_to_learn, 'binary';

Thanks for trying,

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