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From: us on 2 Aug 2010 05:00
"Adam Scott" <sportster360(a)yahoo.com> wrote in message <i35th0$juf$1(a)fred.mathworks.com>...
> and @ the second poster, thanks for a solution, but I'm not sure what you mean by "if I own the symb tbx" and also I wasn't so much looking for another solution, but more a reason mine doesn't work. It is possibly matlab just wont allow you to store all the digits of a number this big and if so I will have to think of another way to do this. But it seems to me like my way SHOULD be able to work, especially since when I tried to look up someone elses code to try to troubleshoot my own, theirs looked very much like mine and returned my same wrong answer.

i thought any solution would be fine and didn't realize you needed help on your particular algorithm...
also: sym tbx = symbolic math toolbox
sorry for confusion...

us
From: John D'Errico on 2 Aug 2010 06:48
"Adam Scott" <sportster360(a)yahoo.com> wrote in message <i35bj8$d6f$1(a)fred.mathworks.com>...
> Ok, sorry if this is a stupid question, I'm still kinda new to Matlab. So I am working on some of the problems from Project Euler and got stuck on question 16 which says
>
>
> "2^(15) = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
> What is the sum of the digits of the number 2^(1000)?"
>
>
> Currently my code for this program is
>
>
> number = 2^1000;
> string = num2str(number);
> i = 1;
> sum = 0;
>
> while i <= 302
> sum = sum + str2num(string(1,i));
> i = i + 1;
> end
>
> fprintf('The sum of the digits of 2^1000 is %f\n', sum)
>
>
> The problem I am having is that when I enter in number = 2^1000 it puts it in scientific notation with only 15 digits after the decimal point and everything after that 0's. I was then really confused when I looked up someone else's matlab code that supposedly got them the right answer, but when I copied it into my matlab it gave me the same wrong answer. The answer I keep getting is 68 while the real answer is 1366. Any help as to what I'm doing wrong? Thanks for your time.
>
> ~Adam~

Without the symbolic TB, you might consider VPI as a
solution. In fact, vpi has helped me to solve a significant
fraction of the 149 problems I've done from PE so far.

sum(digits(vpi(2)^1000))
ans =
1366

You can download it from the file exchange, and it
has the minor virtue of being free.

http://www.mathworks.com/matlabcentral/fileexchange/22725

John
From: Steven_Lord on 3 Aug 2010 10:19


"Adam Scott" <sportster360(a)yahoo.com> wrote in message
news:i35th0$juf$1(a)fred.mathworks.com...
> @ the first poster, I don't know why I made it print as a floating point.
> I guess I got in the habit of printing as a floating point in case it had
> something after the decimal. I don't think that should actually be
> causing any problems though should it? It would just make my output a
> little more sloppy then it should be?
>
> and @ the second poster, thanks for a solution, but I'm not sure what you
> mean by "if I own the symb tbx" and also I wasn't so much looking for
> another solution, but more a reason mine doesn't work. It is possibly
> matlab just wont allow you to store all the digits of a number this big

No. This "problem" is related to IEEE double precision storage.

http://www.mathworks.com/company/newsletters/news_notes/pdf/Fall96Cleve.pdf

When you get up in the 2^1000 range, the spacing between adjacent
representable numbers is greater than 1. In fact, the spacing between the
representation of 2^1000 and the next largest representable number is about
2*10^285.

>> x = 2^1000
x =
1.07150860718627e+301
>> eps(x)
ans =
2.37922705356445e+285

> and if so I will have to think of another way to do this. But it seems to
> me like my way SHOULD be able to work, especially since when I tried to
> look up someone elses code to try to troubleshoot my own, theirs looked
> very much like mine and returned my same wrong answer.

Welcome to the world of floating point arithmetic.

--
Steve Lord
slord(a)mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ
To contact Technical Support use the Contact Us link on
http://www.mathworks.com

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