To determine the remainder of \(8^{2000}\) divided by 7, we apply Fermat's Little Theorem. This theorem states that for a prime number \(p\) and an integer \(a\) not divisible by \(p\), \(a^{p-1} \equiv 1 \pmod{p}\).
In this case, \(a = 8\) and \(p = 7\). Applying Fermat's Little Theorem:
\[8^{7-1} \equiv 1 \pmod{7}\]
\[8^6 \equiv 1 \pmod{7}\]
We can reduce the exponent 2000 modulo 6:
\(2000 \mod 6\) is 2, because:
\(2000 \div 6 = 333\) with a remainder of 2.
Therefore, \(8^{2000} \equiv 8^2 \pmod{7}\).
Calculating \(8^2\):
\(8^2 = 64\).
Finding \(64 \pmod{7}\):
\(64 \div 7 = 9\) with a remainder of \(1\).
Thus, \(8^{2000} \equiv 1 \pmod{7}\).
The remainder when \(8^{2000}\) is divided by 7 is 1.