How many positive integers less than or equal to $6\cdot7\cdot8\cdot9$ solve the system of congruences
\begin{align*}
m &\equiv 5 \pmod{6}, \\
m &\equiv 4 \pmod{7}, \\
m &\equiv 3 \pmod{8}, \\
m &\equiv 3 \pmod{9}?
\end{align*}
So I just solved this step-by-step, I guess — from the first two congruences, we have $6a+5 = 7b + 4$, and dividing by $6$ on both sides (for mod $6$), we have $b + 4\equiv 5 \mod 6$, so $b\equiv 1 \pmod 6$, which gets us $m\equiv 7(1)+4 \mod (6\cdot 7)\equiv 11 \mod 42$.
Then continuing in the same vein I worked on $m\equiv 11 \mod 42$ and $m\equiv 3 \mod 8$: $8c+3 = 42d+11$. Using mod $8$, I have $2d + 3\equiv 3 \mod 8$, which means $d=0$. So $m\equiv 11 \mod 168$ and also $3\mod 9$. Using equations, $168p+11 = 9q+3$. Using mod $9$, this is equal to $6p+2\equiv 3\mod 9$, so $6p\equiv 1 \mod 9$. However, at this point I don't think $p$ has any solutions…
Best Answer
You are correct in your reasoning - so there are no solutions. However, there is a simpler way to realize this:
You know that, if $\gcd(a,b)=1$, the congruences $x\equiv c\bmod a$, $x\equiv d\bmod b$ always have simultaneous solutions. So, you want to look for places where these are not satisfied. This brings you to the congruences with moduli $(6,8)$ and $(6,9)$. Specifically, with the $(6,9)$ case we have
$$m\equiv 5\bmod 6 \implies m\equiv 2\bmod 3$$
as well as
$$m\equiv 3\bmod 9 \implies m\equiv 0\bmod 3.$$
These cannot both be true, and thus there are no solutions.