+2651 Find the number of solutions to
N &\equiv 2 \pmod{6}, \\
N &\equiv 2 \pmod{7}, \\
N &\equiv 2 \pmod{8}
in the interval 0 \le N < 1000.
+1804 For N to satisfy all the choices given, we must find the LCM of 6, 7, and 8.
We have that \(LCM[6,7,8]=168\)
Since they are all containing remainder 2, we have the equation
\(168x+2 \leq 1000\)
Since we want the max number, we set the equations to equal each other and solve for x. We get
\(168x+2=1000\\ 168x=998\\ x=\frac{499}{84} \approx 5.940\)
We round down, so our final answer is 5.
Thanks! :)
