N is a four-digit positive integer. Dividing N by 9 , the remainder is 5. Dividing N by 7, the remainder is 2. Dividing N by 5, the remainder is 4. What is the smallest possible value of N?
+27 Using modular arithmetic, we can create three "equations."
\(n\equiv 5 \pmod{9} \)
\(n\equiv 2 \pmod{7}\)
\(n\equiv4\pmod{5} \)
from the last equation, we get that \(n = 5a+4 \) for some integer a. From the middle one, we get \(n = 7b+2\) for some integer b. Set these equal to each modulo 5 to get that\(5a + 4 \equiv 7b+2 \pmod{5}\). you get that \(2b\equiv 2\pmod{5}\), so \(b \equiv 1 \pmod{5}\). So, \(b = 5c+1\) for some integer c and the bottom system simplifies to 7(5c+1) + 2 = 35c+9, so \(n\equiv 9\pmod{35} \).
We do the same with this and the top equation we left out before to get that \(n = 315e + 149\) for some integer e. e=3 minimizes the four digit integer, and we get n = 1094.
hope this helped 
