about product in uint64

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Nguyen Huy
Nguyen Huy 2021-5-19
编辑： Jan 2021-5-20
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()
  1 个评论
Jan
Jan 2021-5-20
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?

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回答（1 个）

Jan
Jan 2021-5-20
编辑：Jan 2021-5-20
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

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