How to solve Level 45 in Hexologic (easy mode)

Question

For those who don't know how Hexologic works:

You can place one, two or three dots in each blank space and the rows that end in arrows must all add up to the number in the arrow.

Here is Level 45 (easy mode) in Hexologic, which I have been trying to solve for about a week:

_{For those who don't know what those colored cells mean, if I click a cell that is colored cyan, then every other cell that is colored cyan is automatically filled in.}

_{Also, about those greyed out cells with numbers already in them, that's already a part of the level}

Now, the reason I hate this level so much is because there are so many ways that everything can be satisfied. On top of that, there's honestly so much going on that it just hurts to look at.

Is there anything that we can tell is correct right away? Yes, actually:

Here's a brief explanation:

• First of all, there is only one way to satisfy the 13.
• Then, we can notice that since we cannot satisfy the 3 if we put 3 dots in one of the cells, that means that there must be a 2 there, and obviously, we can satisfy the 13, 6, and 3 right away.

Then we can do this because there's only one way to satisfy the 4 and the 8:

And then we can satisfy quite a bit more automatically:

Now this is where I am currently stuck because I don't know what to do from here. This is mostly because now this obviously leads to a bunch of Diophantine equations that I am stuck trying to solve so I can deduce the solution logically. (although I hopefully don't have to do so) So my question is:

How can I solve Level 45 (easy) in Hexologic from here?

Answer 1

Next step is the row with 27 sum, so far the remaining sum is 6.

There are two blue cells here, it can't be 3, since there is still one other cell. It can't be 1, since the other cell will need to be 4. So the blue cell is 2, and the other cell is also 2. Then the rest is forced.

Answer 2

Another way to solve it is to look at the 8 and 4-long 7. The 8 has to be 3-2, and the 4-long 7 2-1-1. So, the intersection needs to be 2. The 11 then needs a 3. Which leaves 4 to be divided 2-2 on the 27 line.

Answer 3

Next, the 6 row on top has a 2, so the other two cells must be 1 and 3, or 2 and 2.

If the 6 row is:
$1 + 3 + 2 = 6$
then the 8 column is:
$8 = 1 + x + 3$
$8 = x + 4$
$4 = x$
which is impossible.

If the 6 row is:
$2 + 2 + 2 = 6$
Then the 8 column is:
$8 = 2 + 3 + 3$
And the 7 column is:
$7 = x + 3 + 3 + 2$
$7 = x + 8$
$-1 = x$
which is impossible.

So it must be:
$3 + 1 + 2 = 6$
$8 = 3 + 2 + 3$
$7 = x + 3 + 2 + 1$

After that, there's a couple of columns with only one missing value. And lastly:

$27 = x + 2 + 3 + 3 + 5 + 2 + 3 + 5 + x$
$27 = 2x + 23$
$4 = 2x$
$2 = x$