FInd an integer m satisfying $\displaystyle n \leq m \leq n+11$ and $\displaystyle m \equiv 7(mod \ 12)$ in each of the following questions.$\displaystyle 10=(-2) \ mod \ 12$ where $\displaystyle -n<r \leq 0$ for $\displaystyle x \equiv r (mod \ n)$.(i) $\displaystyle n=100$

$\displaystyle \Rightarrow 10^2=4 \ mod \ 12$ so I need the number to be $\displaystyle 100+3=103.$

$\displaystyle 10=-2 \ mod \ 12$ where $\displaystyle -n<r \leq 0$ for $\displaystyle x \equiv r (mod \ n)$.(ii). $\displaystyle n=10^6$

$\displaystyle \Rightarrow \ 10^6=(-2)^6 \ mod \ 12 \Rightarrow 10^6=64 \ mod \ 12 \Rightarrow \ 10^6=4 \ mod \ 12$.

But I require a number m such that $\displaystyle m=7 \ mod \ 12.$

Combining my above results reveals the number I need to be $\displaystyle 10^6+3$.

$\displaystyle -9= 3 \ mod \ 12 \Rightarrow \ -9(111)=3(111) \ mod \ 12 \Rightarrow \ -999=9 \ mod \ 12$$\displaystyle (iii). n=-999$

Comparing this with $\displaystyle 7 \ mod \ 12$ gives $\displaystyle m=-999-2=-1001$.

My method seems to be working pretty well (i've checked them all on a calculator) but is there an easier way to do them?