The remainder when 6^1029 is divided by 7 is:

The correct answer is Option (2) → 6

Farmat's Theorem states that if p is a prime and a is not divisible by p:

$a^{p-1}≡1(mod\,p)$

Here, $p=7$ & $a=6$ so,

$6^6≡1(mod\,7)$

$1029÷6=171$ remainder 3

$1029=6×171+3$

$∴6^{1029}=(6^6)^{171}×6^3$

$≡1^{171}×6^3(mod\,7)$

$≡6^3(mod\,7)$

$⇒6^3=216$

$216≡6(mod\,7)$

$∴6^{1029}≡6(mod\,7)$