Let N be the integer of putting all natural numbers from 1 to 35 together
Therefore, N = 1234567891011121314...35
Find the remainder when N is divided by 36
Here's what I have done:
$$N\equiv1(\textrm{mod}\ 2)$$ $$N\equiv0(\textrm{mod}\ 3)$$ $$N\equiv-1(\textrm{mod}\ 4)$$ $$N\equiv0(\textrm{mod}\ 9)$$
We are finding $N (\textrm{mod}\ 36)$, so how do we put all those congruences up there together and find the answer?