From: United States
First, a minor piece.
You can eliminate the 2's from the last row, columns 5 and 6. No matter which way they go, the 2 will have to be in either column 1 or 3.
Now, for something better.
The top of column 3 is 26 down. The bottom 2 digits cannot be 3 and 5. Not just because it would make it impossible to determine the order between them and the 3-5 in column 2 (which I do not accept as valid logic for an elimination). But if you put in the 3 and 5, in either order, that puts a 1 at the third position. Tracing that out, you can fill in a ton of numbers, but you eventually end up with row 5, 33 across, having a 1, 2, 3, 4 and 5. There is no way to make the 33 with all of those. Therefore you can eliminate the 3 and 5 as possibles for the bottom 2 positions of the 26 down in column 3.
That leaves you with either 2-4 or 2-6 and whichever it is, the 2 must be the bottom number.
Knowing that the 4th position must be 4 or 6 eliminates the 3 as a possible starting number for row 4, 32 across.
Hope that helps.