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Nguyen Huy on 19 May 2021
Edited: Jan on 20 May 2021
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 on 20 May 2021
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?

Jan on 20 May 2021
Edited: Jan on 20 May 2021
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