# Why this function infinity loop

3 views (last 30 days)
Nguyen Huy on 7 May 2021
Commented: Walter Roberson on 8 May 2021
%It give 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. Example x=2 so n=5 because 2^5=32 and last
%digits of 32 is 2
%So why this take infinity loop
function n = bigNumRepeat(x)
d = length(num2str(x))
y=2
n=2
while 1
y = y*x
y = num2str(y)
y = str2num(y(end - d + 1:end))
if y==x
break
end
n=n+1
end
end

John D'Errico on 7 May 2021
Because if your powers grow larger than 2^53-1 (i.e., flintmax), the number is no longer expressable exactly as an integer. All of those operations with str2num will produce garbage at some point. You CANNOT do this using doubles if the number grows too large.
Walter Roberson on 8 May 2021
x = randi([11 99])
x = 93
d = length(num2str(x))
d = 2
y = x;
n = 1;
flag = true;
modulus = 10^d;
while flag && n <= 1000
n = n + 1;
y = mod(x*y,modulus);
flag = y ~= x;
end
if flag
else
n
end
n = 5
Note that x is not relatively prime with 10^d then there might not be a solution.

Walter Roberson on 7 May 2021
Edited: Walter Roberson on 7 May 2021
y=2
y = y*x
That does not give you x^n, it gives you 2 * x^n . You should initialize y = 1
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 7 May 2021
then initialize y=x