Find the least residue modulo 7 of the quantity $(22 \times 51 ) + 698^5$.
What I did was as follows:
I know $22 \equiv 1 $ (mod 7), $51\equiv 2$ (mod 7) and that $698 \equiv 5\equiv -2$ (mod 7).
So,
$22 \times 51 + 698^5 \equiv (1 \times 2) + (-2)^5\equiv -30 \equiv -2$ mod 7.
Thus the least residue modulo 7 is -2.
I want to know if my thinking process is correct and also can the least residue modulo be 5?
Yes it is correct and
$$(22 \times 51) + 698^5\equiv-2 \pmod 7$$
$$(22 \times 51) + 698^5\equiv 5 \pmod 7$$
are equivalent.
In some cases, for convenience, we can fix and use $0,1,2,3,4,5,6$ as residue classes $\pmod 7$ in other cases we can fix $-3,-2,-1,0,1,2,3$ but they are obviously equivalent choices.