Highest power of 2 that divides n.

Never mind I messed up.

n=sym('94315998522576010519588224930693232398146802027362761139521');
a=7;
i=1;%this is a counter that starts from 1.

for i=1:inf
    remainder=mod(n-1,2^i);
    if remainder~=0
        break
    end
end

r=i-1%r takes the second last value of i because 2 raised to the last value is not divisible by n-1
s=rdivide(n-1,2^r)%this is the integer we get after n-1 is divided by the 2^(i-1)

[David Hill]

I am very confused. If you factor n-1, you get:

[ 2, 2, 2, 2, 2, 2, 3, 3, 3, 5, 13, 64763, 36377857, 478202419, 745336575801888629040192801767759];

How is a power of two ever going to be divisible by n-1? I obviously don't understand what you are trying to do.

[Yuechuan Chen]

Yeah I noticed what I did was very wrong. I was basically trying to set up a loop so that n-1 divides 2 repeatedly until the remainer is not 0.

I tried doing this

while mod(n-1,sym(2)^i)==0
    i=i+1;
end

which also didn't get me anywhere cloe to the correct answer.

How would i get around this question? Thanks!

Answer 1

[James Tursa]

In addition to David's comments, you need to do all of the calculations symbolically, so all of these should be sym: 2, i, r.

[Yuechuan Chen]

Oh I see, thank you. that why it gave me some stupid numbers. I'm trying to find a way to divide n-1 by 2 repeatedly until the remainder isn't 0 and output how many times it has iterated which would be the answer I want, but I've been trying out all sorts of stuff but it didn't quite work out.

Answer 2

[Yuechuan Chen]

n=sym('94315998522576010519588224930693232398146802027362761139521');
a=7;
i=1;
while mod(n-1,2^i)==0
    i=i+1;
end
i
r=i-1%r takes the second last value of i because 2 raised to the last value is not divisible by n-1
s=vpa((n-1)/2^r)

Nevermind I think I've got it, My question was very unclear very sorry about that.