How to find the least $N$ such that $N \equiv 7 \mod 180$ or $N \equiv 7 \mod 144$ but $N \equiv 1 \mod 7$? How to approach this problem:

N is the least number such that $N \equiv 7 \mod 180$  or $N \equiv 7 \mod 144$  but $N \equiv 1 \mod 7$.Then which of the these is true:



*

*$0 \lt N \lt 1000$

*$1000 \lt N \lt 2000$

*$2000 \lt N \lt 4000$

*$N \gt 4000$


Please explain your idea.
ADDED: The actual problem which comes in my paper is "or" and the "and" was my mistake but I think I learned something new owing to that.Thanks all for being patient,and appologies for the inconvenience.

(1) For the original version of the question $\rm\:mod\ 180 \ $ and $\rm\: mod\ 144\::$
$\rm\: 144,\:180\ |\ N-7\ \Rightarrow\ 720 = lcm(144,180)\ |\ N-7\:.\:$  
So, $\rm\: mod\ 7:\ 1\equiv N = 7 + 720\ k\ \equiv -k\:,\:$ so $\rm\:k\equiv -1\equiv 6\:.$
Thus $\rm\: N = 7 + 720\ (6 + 7\ j) =\: 4327 + 5040\ j\:,\:$ so $\rm\ N\ge0\ \Rightarrow\ N \ge 4327\:.$
(2) For the updated simpler version $\rm\:mod\ 180\ $ or $\rm\  mod\ 144\:,\:$ the same method shows that   
$\rm\: N = 7 + 180\ (3+ 7\ j)\:$ or $\rm\:N = 7 + 144\ (2 + 7\ j)\:,\:$ so the least$\rm\ N> 0\:$ is $\rm\:7 + 144\cdot 2 = 295\:.$
SIMPLER $\rm\ N = 7+144\ k\equiv 4\ k\ (mod\ 7)\:$ assumes every value $\rm\:mod\ 7\:$ for  $\rm\:k = 0,1,2,\:\cdots,6\:,\:$ and all these values satisfy $\rm\:0 < N < 1000\:.\:$ Presumably this is the intended "quick" solution.
