What is the remainder when 6^17 + 117^6 is divided by 7?
How to approach these type of questions?
What is the remainder when 6^17 + 117^6 is divided by 7?
How to approach these type of questions?
On
$6^{17} = (7-1)^{17} = (-1)^{17} + \text{multiples of }7$ $117^6 = (7\times 17 - 2)^6 = (-2)^6 + \text{multiples of }7 $ sum up the two quantities, you get$$ 6^{17}+ 117^6 = 63 + \text{a multiple of }7 = \text{a multiple of }7 $$ so the reminder is 0.
On
Use arithmetic $\pmod 7$. Specifically we'll use these facts:
$x^n \pmod y \equiv (x \pmod y)^n \pmod y$
and
$(y - 1) \pmod y \equiv (-1) \pmod y$
Using those facts...
$6^{17} + 117^6 \pmod 7 \equiv$
$(-1)^{17} + (5)^6 \pmod 7$
$5^2 \equiv 4 \pmod 7$ and $5^4 \equiv 4^2 \equiv 2 \pmod 7$, so
$-1 + (4 \times 2) \equiv 0 \pmod 7$
It helps a lot to know about modular arithmetic. Then you see that
$$6\equiv-1\mod 7\qquad\text{and}\qquad117\equiv-2\mod7$$
and thus
$$6^{17}+117^6\equiv(-1)^{17}+(-2)^6=-1+64=63\equiv0\mod7$$
It helps even more to know about Fermat's little theorem, because then you would know that $117^6\equiv1\mod7$ directly (as soon as you check that $7$ doesn't divide $117$).