(This answer is to show the logic after the initial 'simple' first 25 digits, after which it gets much harder)
So, we have our initial 25 digits, as shown in @Amorydai's answer. The grid looks as follows:
Now the next step of logic, as indicated by the hint, involves the 10 sandwich in the middle column:
The 10 can be made up of two pairs, either 4/6 or 8/2. Lets see what happens if we make it a 4/6:
The 4 leaves only one place a 4 can go in C8. This 4 removes a candidate in the same box in R7, leaving the only 4 in row 7 as the highlighted red cell.
However, now take a look at R8. Every single free cell in R8 is either in the same column, or the same box as a 4, leaving no place for a 4 in R8. Therefore, the 10 pair MUST be the 8/2.
With a bit of normal sudoku logic we get to this point:
The 6 in R7 could only go in one place, as could the 6 in R8. There is now 3/4s in the bottom two rows in 2 boxes, meaning the bottom left box has a 258 triple, and the 8 in C8 must be at the bottom. There are a couple of 3/4 pairs, and some row restrictions towards the bottom.
Now looking at the 11 pair in the left column:
There are two options, it could be 7/4 or 3/8 (or 8/3). Lets see what happens if it's 7/4:
The cells highlighted in yellow are affected and the 3s and 4s in the bottom rows are mostly solved.
However, the interesting bit is C1 and C2. Both have 3 numbers remaining, 2/5/8, highlighted in blue. The problem is the red cell, no matter which number it is, it covers all 3 blue cells in the box below, meaning whatever the red cell is, the box below would not have a place for that number. Hence the 11 pair cannot be 7/4 and is 3/8 or 8/3. And as there is a 2/5/8 triple in the bottom left box, it must be 8/3.
Using these deductions, and with some sudoku logic the digits start to fall into place:
The 3 has a domino effect for the 3s and 4s towards the bottom. The right hand column can be solved, as can the 6s towards the top few rows. The logic gets easier from this point as more numbers are placed.
From here the puzzle can be solved fairly easily:
By going around multiple rows or columns can be solved, and many pairs are established. The bottom right box can be solved now there is a 4 in C6, and solving C6 will pretty much give us everything needed to finish off the puzzle.
Solving with normal sudoku logic from here leads us to the answer: