Thread: Number theory

Hi i need help with this problem.. The quest is what positive rest do i get if i divide it with 8
7^41+17^20-6



I should post this on number Theory section! Im sorry

Let $\displaystyle x = 7^{41}+17^{20}-6$. Determine $\displaystyle x \mod 8$ ("positive rest" I assume means positive remainder).

$\displaystyle 7 \equiv -1 \mod 8$, so $\displaystyle 7^{41} \equiv (-1)^{41} \equiv -1 \mod 8$.

$\displaystyle 17 \equiv 1 \mod 8$, so $\displaystyle 17^{20} \equiv 1^{20} \equiv 1 \mod 8$.

Thus $\displaystyle x \equiv 7^{41}+17^{20}-6 \equiv (-1) + (1) - 6 \equiv -6 \equiv 2 \mod 8$.

Thus x leaves remainder 2 when divided by 8.