HP Forum Archive 17

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Upcoming 35s question #2 Message #1 Posted by Gene Wright on 11 July 2007, 10:58 p.m. The answer to question 1 was E. The early models did not have a lowered-ON key. Question 2: In the spirit of guessing, how many full-precision real numbers can be stored in registers on the HP 35s? No multiple choice this time.

Re: Upcoming 35s question #2 Message #2 Posted by Thomas Radtke on 12 July 2007, 4:08 a.m., in response to message #1 by Gene Wright Let me say, as much as can be represented by floating point numbers within the limits of the machine :^). (I was right on q#1, I can afford to fail this time :-) Edited: 12 July 2007, 7:19 a.m.

Re: Upcoming 35s question #2 Message #3 Posted by Gene wright on 12 July 2007, 9:26 a.m., in response to message #2 by Thomas Radtke Answer is well over 2400 floating point, full precision numbers.

Re: Upcoming 35s question #2 Message #4 Posted by Paul Dale on 12 July 2007, 4:47 p.m., in response to message #3 by Gene wright That isn't exactly a precise answer to the question... - Pauli

Re: Upcoming 35s question #2 Message #5 Posted by Gene Wright on 12 July 2007, 5:24 p.m., in response to message #4 by Paul Dale Lol. Ok, 2514 is the actual, non-useful number. 801 x 3 + 26 x 3 + 6 x 3 + 5 x 3 which fills up all 801 indirect registers, all 26 lettered registers, all 6 stat registers (which have to be stored indirectly) and all 4 stack registers and lastx too. Reality is that you can now have indirect registers 1 through 800 filled with 3-D vectors containing 3 full precision real numbers using the program supplied in the learning module. That ought to do for now. :-)

Re: Upcoming 35s question #2 Message #6 Posted by Thomas Radtke on 12 July 2007, 5:24 p.m., in response to message #4 by Paul Dale Indeed, my answer was way better. You just have to read the question the right way ;-).

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