Solving $2^k +k \equiv 0 \pmod {323}$ Find all $k$ such that
$$2^k + k \equiv 0 \pmod{323}.$$

I noticed that $323 = 17\cdot 19$ so I thought about using the Chinese Remainder theorem by considering $2^k+k$ modulo 17 and 19. I got $k \equiv 16 \pmod{17}$ and $k\equiv 18 \pmod{19}$, which gives $k\equiv 322 \pmod{323}$. However, after trying this, the given solution was not valid : $2^{322} + 322 \equiv 156 \not \equiv 0 \mod{323}$.
Can someone explain why this didn't work, and what I can do to solve this problem? Thanks.
 A: As explained in the comments, it didn't work because CRT wasn't applied correctly.
In particular, the solution to $ 2^k + k \equiv 0 \pmod{17}$ has a cycle length of $ 16 \times 17$, because $2^k$ has a cycle length of 16 and $k$ has a cycle length of 17.
So, the (theoretical) approach is to find all the solutions to $2^k+k \equiv 0 \pmod{17} $ working in $\pmod{17 \times 16}$, and likewise for the other equation working in $\pmod{19 \times 18}$, and finally combine them via CRT in $ \pmod{17 \times 19 \times 144}$.
If you understand the above logic, I do not recommend actually finding all the solutions, as it's just a very tedious process that's best left to the computer.
A: Okay to solve $2^k \equiv \pmod {17}$ we know by FLT than $2^{16}\equiv 1 \pmod {17}$ and as $2^4 =16 \equiv -1$ that $2^8\equiv 1 \pmod {17}$.
So if $k \equiv 0, 1,2,3,4....,7\pmod {16}$ we have $2^k\equiv 1,2,4,8,-1,-2,-4, -8\pmod {17}$ and by CRT will be a unique solution for each pair. $k\equiv a \pmod {16}$ and $k \equiv 2^a \pmod {16}$. So that is eight solutions $\pmod {16*17}$.
(exmaple:  If $k\equiv 0\pmod {16}$ and $k \equiv 2^0=1\pmod {17}$ then $k\equiv -16 \equiv 256 \pmod {272}$.  Or if $k \equiv 1 \pmod 16$ and $k \equiv 2^1 \equiv 2\pmod {17}$ the solution $\pmod{272}$ is .... whatever)
Then we can do the same thing for $k\equiv b\pmod {18}$ and $k\equiv 2^b\pmod {19}$
So there are $18$ solutions $\pmod{18\cdot 19}$ for those.
However as $\gcd(16,18)=2\ne 1$ we must have $a,b$ either both even or both odd.
Still that is $4$ solutions $\pmod {16*17}$ with $a$ odd, and $4$ \solutions $\pmod{16*17}$ with $a$ even.  And $9$ solutions $\pmod{18*19}$ with $b$ even or odd.
So that is  $4*9=36$ solutions $\pmod{\frac {16*17*18*19}2}$ for $a,b$ odd and $36$ for $a,b$ even.
So there are $72$ solutions.
And it's way to much work do to.
But to find one solution we can have $k\equiv 256 \pmod {272}$
And $k \equiv 0 \pmod {18}$ and $k\equiv 2^0=1 \pmod{19}$ so $k\equiv -18\equiv 18^2 \mod {18*18}$ or $k \equiv 324 \pmod {342}$.
So if we solve $k\equiv 256\pmod {272}$ and $k \equiv 324{342}$ that solution ought to give us $2^k \equiv k\pmod {323}$ and ... oh, that $2^k - k\equiv 0$.... oh well the same ideaa will hold.
