>>11003883I’m not sure I’m seeing what you’re seeing.
There are :
Z/144Z = Z/16Z x Z/9Z
Z/16Z x Z/3Z2
Z/8Z x Z/2Z x Z/9Z
Z/8Z x Z/2Z x Z/3Z2
Z/4Z2 x Z/9Z
Z/4Z2 x Z/3Z2
Z/4Z x Z/2Z2 x Z/9Z
Z/4Z x Z/2Z2 x Z/3Z2
Z/2Z4 x Z/9Z
Z/2Z4 x Z/3Z2
If p is neither 2 nor 3, (p^l)G and (p^(l+1))G are equal to the group itself so the answer is always 0. So please don't ask.
For p^l=3, it is either 0 (Z/3Z2) or 1 (Z/9Z). This is the only way you can know the 3-part so you HAVE to ask this question (or the equivalent p=3, l=0 which gives 1 or 2). Now you only have one question left to know the 2-part.
For p^l=2, the answer is the number of factors in the 2-part of G which are not Z/2Z (it can be 0 if Z/2Z4, 1 if Z/16Z or Z/8ZxZ/2Z or Z/4ZxZ/2Z2, 2 if Z/4Z2). So it is not enough (your winning probability is 2/5+3/5*1/3=3/5=60%).
For p=2 l=0 : the answer is the number of factors of the 2-part of G (1 if Z/16Z, 2 if Z/8ZxZ/2Z or Z/4Z2, 3 if Z/2Z2xZ/4Z, 4 if Z/2Z4). This is also not enough, but this time you have 3/5+2/5*1/2=4/5=80% chances of winning.
For p^l=4, the answer (numbers of 2-factors that are not Z/2Z or Z/4Z) can be 0 (Z/2Z4 or Z/2Z2xZ/4Z or Z/4Z2) or 1 (Z/16Z or Z/8ZxZ/2Z). This time you have 3/5*1/3+2/5*1/2=2/5=40% chances of winning.
For p^l=8, the answer can be 1 (Z/16Z) or 0 (everything else). This means 1/5+4/5*1/4=2/5=40% chances of winning too.
For p^l=16,32,64... the answer is always 0. So you have 20% chances of winning.
This whole thing proves it's impossible but you have 80% chances of survival if you're smart, and you can't do better.
If you have three questions, just ask 31, 2^0 and the 21 one (or even 22). You will always survive.
(Sauce: the comment section lol)
