The smallest such number is 1055.
We want to find [tex]x[/tex] such that
[tex]\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}[/tex]
The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.
[tex]x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}[/tex]
[tex]x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}[/tex]
[tex]x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}[/tex]
Taking everything together, we end up with the system
[tex]\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}[/tex]
Now the moduli are coprime and we can apply the CRT.
We start with
[tex]x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5[/tex]
Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.
Taken modulo 2, we end up with
[tex]x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2[/tex]
which means the first term is fine and doesn't require adjustment.
Taken modulo 3, we have
[tex]x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3[/tex]
We want a remainder of 2, so we just need to multiply the second term by 2.
Taken modulo 5, we have
[tex]x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5[/tex]
We want a remainder of 0, so we can just multiply this term by 0.
Taken modulo 7, we have
[tex]x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7[/tex]
We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since [tex]2\cdot4\equiv8\equiv1\pmod7[/tex], the inverse of 2 is 4.
So, we have to adjust [tex]x[/tex] to
[tex]x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845[/tex]
and from the CRT we find
[tex]x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}[/tex]
so that the general solution [tex]x=210n+5[/tex] for all integers [tex]n[/tex].
We want a 4 digit solution, so we want
[tex]210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5[/tex]
which gives [tex]x=210\cdot5+5=1055[/tex].
