What is the remainder when the product of the primes between 1 and 100 is divided by 16? The product of all the prime numbers between 1 and 100 is equal to $P$. What is the remainder when $P$ is divided by 16?
I have no idea how to solve this, any answers?

Skip the first prime $2$ and look for the product modulo $8$.
The twenty-four odd primes $<100$ are
$$3,5,7,11,13,17,  19,23,29,31,37,41,  43,47,53,59,61,67, 71,73,79,83,89,97. $$
Modulo $8$, these are
$$3,5,7,3,5,1,3,7,5,7,5,1,3,7,5,3,5,3,7,1,7,3,1,1$$
so their product is $$1^53^75^67^6\equiv 3\pmod 8 $$
(where we might profit from using $x^2\equiv 1\pmod 8$ for odd $x$).
Thus $P\equiv 6\pmod{16}$.
