about product in uint64

My question is iven an integer x which contains d digits, find the value of n (n > 1) such that the last d digits of x^n is equal to x. If the last d digits will never equal x, return inf.

Example :

%x = 2; (therefore d = 1)
%2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32
%n = 5;

So my solution is take x into uint64 vecto,then take the x^n by take product of a*x with example is

a=uint64([x])
for i=1:inf
a=a*x
for j=length(a):-1:1
     b=rem(a(j)/10)
     c=mod(a(j),10)
     if a(j)>10
     a(j)=c
     a(j-1)=a(j-1)+b
end
end
if x==a(end-d:end)
break
end
end

%example : 0 0 0 2 *

2

==========

0 0 0 4 => [0 0 0 4]*2 =>[0 0 0 8]*2=>[0 0 0 16]=[0 0 1 6]*2=>[0 0 0 32]=[0 0 3 2] have a(end-d:end)==x then break

so my question is how could i product the number with over 2 digits in uint64()

[Jan]

Why do you use uint64 instead of uint8 to store decimal digits?

What about limiting the solution to the UINT64 range, which is at least up to 1.8e19?

Answer 1

[Jan]

With limiting the values to UINT64 range:

x   = uint64(2);
d10 = uint64(10^(floor(log10(x)) + 1));

n   = 2;
a   = x^n;
lim = intmax('uint64');
while rem(a, d10) ~= x && a <= lim
   a = a * x;
   n = n + 1;
end

if a < lim
    disp(n)
else
    disp('search failed.');
end