Number Theory

By Chinese Remainder Theorem, there exists a unique residue $\pmod{315}$ for $N$. Hence, all we need to do is find a possible residue, and that will be the unique residue for $N$ when taken $\pmod{315}$. By $N \equiv 3 \equiv -4 \pmod{7}$ and $N \equiv 5 \equiv -4 \pmod{9}$, we get that $N \equiv -4 \pmod{63}$. Hence, $N = 63k + 59$ for some integer $k$.

Combining this with the last condition, we get $63k + 59 \equiv 3k + 4 \equiv 4 \pmod{5}$, so $3k \equiv 0 \pmod{5}$. Multiplying both sides of this congruence by $2$, we get that $6k \equiv k \equiv 0 \pmod{5}$, so hence $k \equiv 0 \pmod{5}$. Thus, $N \equiv 63 \cdot 0 + 59 \equiv 59 \pmod{315}$. 

Hence, we want to find the smallest four-digit $N$ where $N \equiv 59 \pmod{315}$, which gives us an answer of $315 \cdot 3 + 59 = \boxed{1004}$.

LCM(9,7,5) ==315

LCM(3, 4, 5)==60

315n  + (60 - 1)==315n  +  59, where n==0, 1, 2, 3,4,........etc.

The smallest 4-digit integer ==[314 x 3  +  59] ==945  +  59 ==1,004