Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively. 
Find an integer having the remainders $2,3,4,5$ when divided by $3,4,5,6$, respectively.

My work:
We consider the congruences $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$, $x \equiv 5 \pmod 6$. We can reduce this further to $x \equiv 2 \pmod 3$, $x \equiv 3 \pmod 4$, $x \equiv 4 \pmod 5$. We have
$N_1 = 4 \cdot 5 = 20 \implies 20 x_1 \equiv 1 \pmod{3} \implies 2x_1 \equiv 1 \pmod{3} \implies x_1 \equiv 2 \pmod {3}$
$N_2 = 3 \cdot 5 = 15 \implies 15x_2 \equiv 1 \pmod{4} \implies -x_2 \equiv 1 \pmod{4} \implies x_2 \equiv 3 \pmod {4}$ 
$N_3 = 3 \cdot 4 = 12 \implies 12 x_3 \equiv 1 \pmod{5} \implies 2x_3 \equiv 1 \pmod{5} \implies x_3 \equiv 3 \pmod {5}$
Now, 
\begin{align*}
\bar x &= a_1 N_1 x_1 + a_2 N_2 x_2 + a_3 N_3 x_3 \\
&= 3 \cdot 20 \cdot 2 + 4 \cdot 15 \cdot 3 + 5 \cdot 12 \cdot 3 \\
&= 480 \equiv 0 \pmod {3 \cdot 4 \cdot 5}
\end{align*}
Is this correct, or is something wrong in my work? I don't like how I have $0$ remainder.

I too found the answer to be: x ≡ -1 (mod 60).
My solution:

