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.
 A: (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.
A: The title says "or" and the text says "and". I will assume "and".
We want $N$ to be congruent to $7$ modulo $180$ and modulo $144$.  This will be true iff $N$ is congruent to $7$ modulo the LCM of $180$ and $144$, which is $720$.
So $N$ must have shape $N=720k+7$ for some integer $k$.
But we want $N \equiv 1 \pmod{7}$. 
Since $N=700k +20k +7$, we can see that $N\equiv 20k \pmod{7}$.
Presumably we want $N$ positive, though this was not specified. It is easy to see that the least positive $k$ that works is $k=6$.  Why is it so easy? Note that $20\equiv -1 \pmod{7}$.  So to make $20k \equiv 1 \pmod{7}$, we must have $k \equiv -1\pmod{7}$.  The least positive $k$ congruent to $-1$ is $6$.
That forces $N>4000$.
Added: The text of the original question said $180$ and $144$. 
For the "or" version, we note that $N \equiv 7 \pmod{\gcd(180,140)}$.
Thus $N\equiv 7 \pmod{36}$, or equivalently $N$ is of the shape $36k+7$.  In particular, since $N \equiv 1 \pmod 7$, we must have $k\equiv 1 \pmod 7$. Probably at this stage (or earlier!) search is most efficient. Try $k=8$. That gives $N=295$, which works, since $295=(2)(144)+7$.
A: I assume that the question is "or."  Then it is simplest to just compute the answer.
Notice that $180 \equiv 5 \pmod{7}$ and $144\equiv 4 \pmod{7}$.  Since $3\cdot 5\equiv 1\pmod 7$, we see that $3\cdot 180+7\equiv 1\pmod{7}$.  Also, since $2\cdot4\equiv 1\pmod{7}$ we also have $2\cdot144 +7\equiv 1\pmod{7}$.  As the smaller of the two is the second, we conclude that $N=295$.
