To find the product \( P \) of all the prime numbers between 1 and 100 and then determine the remainder when \( P \) is divided by 16, we first list all the prime numbers within the specified range:
The prime numbers between 1 and 100 are:
\[ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. \]
Let's denote \( P = 2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29 \times 31 \times 37 \times 41 \times 43 \times 47 \times 53 \times 59 \times 61 \times 67 \times 71 \times 73 \times 79 \times 83 \times 89 \times 97 \).
Next, we analyze the product \( P \) modulo 16. Since 16 is a power of 2, and \( P \) includes the factor 2 (the only even prime), it is crucial to count the powers of 2 present in \( P \).
The prime factorization of \( P \) has \( 2 \) included, so we start by counting how many times 2 appears in the list of primes:
1. The prime \( 2 \) contributes one \( 2 \) to \( P \).
Next, observe that powers of 2 beyond 2 will affect the final product. However, since \( P \) contains other odd primes, we can focus on the contribution of the prime \( 2 \). The product of the odd primes will not introduce any more factors of 2.
The remaining primes contributing to \( P \) are all odd primes. Thus, the total contribution of the factor of 2 in \( P \) is simply 1 (from the prime \( 2 \)).
Next, we calculate \( P \) modulo 16. Since \( 2^1 = 2 \), and the only factor we observe is the \( 2 \) from the prime \( 2 \), we need to determine how this stacks up against modulo 16.
Thus, if \( 2 \) is multiplied by any odd integer (which will not contribute any additional factors of \( 2 \)), the total multiplicative contribution leaves us with:
\[
P \equiv 0 \mod 2 \quad \text{(because \( P \) contains \( 2 \))}
\]
However, we need to also check higher powers, specifically modulo 16:
Given our observations, we know all primes greater than 2 are odd, and they contribute no further factors of 2. Hence their contribution remains odd. Thus:
- We can deduce \( P \) will have \( 1 \) factor of \( 2 \) and will result in being odd. Final multiplicative contributions through odd primes keep odd.
So effectively:
\[
P \equiv 2 \mod 16.
\]
Upon gauging the total contribution of primes and realizing how only \( 2^1 \) could factor in, squaring this with non-even factors leads us to return:
Thus, the final modulus we seek:
\[
\text{The remainder when } P \text{ is divided by } 16 \text{ is } \boxed{0}.
\]
Therefore correct formulation yields the answer \( 0 \) based on the attained even multipliers via primal impact over non-even prime assemblies.
