How to increase precision in this case?

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Daniel
Daniel 2022-10-19
编辑： Daniel 2022-10-19
I have a code which finds the first 10 consecutive numbers after decimal that forms a prime number:
k = 10;
while true
a = mod(floor((pi-3)*10^k),10^10);
if isprime(a) == 1
disp(a)
break
endif
k += 1;
endwhile
And this only works for pi, I also need it to e but its not enough precision
What can i do here?

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John D'Errico
John D'Errico 2022-10-19
编辑：John D'Errico 2022-10-19
Ran in:
Take it from here:
vpa(exp(sym(1)),500)
ans = 
2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174135966290435729003342952605956307381323286279434907632338298807531952510190115738341879307021540891499348841675092447614606680822648001684774118537423454424371075390777449920695517027618386062613313845830007520449338265602976067371132007093287091274437470472306969772093101416928368190255151086574637721112523897844250569536967707854499699679468644549059879316368892300987931
Now just process the digits.
Or I think I have something on the order of a million digits of e and pi stored in my HPF toolbox. (Might be only a half million.)
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John D'Errico
John D'Errico 2022-10-19
Ran in:
Can you extract the string of digits? I think you are fixated on the idea that you NEED to work with numbers, and there is only one possible way to extract those digits. That is completely wrong. I showed you how to extract all the digits you want. Perhaps I stopped one step short, thinking you could take it from there. So I'll add a next step.
D = char(vpa(exp(sym(1)),500))
D = '2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174135966290435729003342952605956307381323286279434907632338298807531952510190115738341879307021540891499348841675092447614606680822648001684774118537423454424371075390777449920695517027618386062613313845830007520449338265602976067371132007093287091274437470472306969772093101416928368190255151086574637721112523897844250569536967707854499699679468644549059879316368892300987931'
That is not a string of numeric digits there, but a string of characters. But even so, it is all you need. Feel free to delete the decimal point even.
D(2) = []
D = '27182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174135966290435729003342952605956307381323286279434907632338298807531952510190115738341879307021540891499348841675092447614606680822648001684774118537423454424371075390777449920695517027618386062613313845830007520449338265602976067371132007093287091274437470472306969772093101416928368190255151086574637721112523897844250569536967707854499699679468644549059879316368892300987931'
You can even convert them into actual numeric digits if you feel the need, but you need to make some effort, and I'm sorry, but I won't do your homework, and I've already done almost all of it here.
Daniel
Daniel 2022-10-19
编辑：Daniel 2022-10-19
Thank you John. We learned only loops (for, while) on the last lesson. I didnt know about it

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