Can someone provide me a simplest solution for this?

[Star Strider]

Post your solution to it first.

[Koo Ian Keen]

close all; clear all; clc;

%Takes about 20 seconds to run this code

iter = 0;

for n = 1:50000 %Change the 50000 to whatever values ur question wants

nstr = num2str(n);

nlength = length(nstr);

N = n^3;

Nstr = num2str(N);

Nlength = length(Nstr);

x = 0;

for i = 1:nlength

x = x + str2num(Nstr(Nlength - (i - 1)))*(10^(i - 1));

if x == n

iter = iter + 1;

else

continue

end

end

end

fprintf('The answer is %d', iter)

so this is the solution i got from my course mate, i dont understand the ' x = x + str2num(Nstr(Nlength - (i - 1)))*(10^(i - 1));' part

Answer 1

[John D'Errico]

Easy peasy? Pretty fast too.

tic
n = 1:50000;
M = 10.^(floor(log10(n))+1);
nsteady3 = sum(n == mod(n.^3,M));
toc

Elapsed time is 0.010093 seconds.

nsteady3
nsteady3 =
    40

So it looks like 40 of them that do not exceed 50000. Of course, 1 is a steady-p number for all values of p. The same applies to 5, and 25.

If n was a bit larger, things get nasty, since 50000^3 is getting close to 2^53, thus the limits of double precision arithmetic to represent integers exactly. And of course, if p were large, that would restrict things too. I'll leave it to you to figure out how to solve this for a significantly larger problem, perhaps to count the number of solutions for p=99, and n<=1e15. (Actually, this is quite easy as long as the limit on n is not too large. So n<=5e7 is quite easy even for p that large, and it should be blazingly fast.)

I got 78 such steady-99 numbers that do not exceed 5e7. It took slightly longer, but not too bad.

Elapsed time is 5.547148 seconds.