TITLE: What is the remainder when $16^{15}-8^{15}-4^{15}-2^{15}-1$ is divided by $96$?
QUESTION [5 upvotes]: I found this question from an old math questionnaire.
What is the remainder when $$16^{15}-8^{15}-4^{15}-2^{15}-1$$ is divided by 96?
I already know the answer. It is 31, says the answer at the back of the questionnaire. I just do not know why 31 is the answer. I do not know how the process of getting the remainder is done.
I tried a lot of crazy things and got answers like -135 and -63/32 which are obviously wrong, so I would not tell anymore what I did. Can anyone help me?

REPLY [7 votes]: Since you don't know modular arithmetic we can instead give a direct divisibility proof.
$$2^{60}\!-2^{45}\!-2^{30}\!-2^{15}\!-1\, =\, \underbrace{2^{30}(2^{30}\!-1)-2^{5}(2^{40}\!-1)-2^{5}(2^{10}\!-1)}_{\Large\color{#c00}{ 96}\, n}\!-2^5\!-2^5\!-1$$
$\color{#c00}{96} = 2^5\cdot 3\,$ divides all RHS summands of form $\,2^J (2^{2K}\!-1)$ since  $\,J\ge 5\,$ so $\,2^5$ divides $2^J,\,$ and $\,3\,$ divides $\,2^{2K}\!-1\ $ (put $\ a=2\ $ in  $\,a+1\,$ divides $\,a^2-1\,$ divides $\,(a^2)^K-1).$
So it has form $\,\color{#c00}{96}\,n - 2^5\!-2^5\!-1 = 96n-65 = 96(n\!-\!1)+96-65  = 96(n\!-\!1)+\color{#0a0}{31}$