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To determine the remainder of \(6^{1029}\) when divided by 7, we utilize Fermat's Little Theorem. This theorem states that for a prime number \(p\) and an integer \(a\) not divisible by \(p\), the congruence \(a^{p-1} \equiv 1 \pmod{p}\) holds.
In this case, \(p = 7\) and \(a = 6\). Applying Fermat's Little Theorem, we get \(6^{7-1} = 6^6 \equiv 1 \pmod{7}\).
The exponent \(1029\) needs to be expressed in relation to the exponent in the theorem (which is 6). We divide \(1029\) by 6: \(1029 = 6 \times 171 + 3\).
Consequently, \(6^{1029} = 6^{6 \times 171 + 3} = (6^6)^{171} \times 6^3\). According to Fermat's Theorem, \((6^6)^{171} \equiv 1^{171} = 1 \pmod{7}\).
Next, we evaluate \(6^3 \pmod{7}\):
\(6^3 = 216\). To find \(216 \mod 7\):
Dividing \(216\) by \(7\) yields \(30\) with a remainder of \(6\), so \(6^3 \equiv 6 \pmod{7}\).
Therefore, the remainder when \(6^{1029}\) is divided by 7 is 6.