STEP 3

LET’S start putting in ‘twins’ in the middle-right box. Look at the cage with a total of 8. The only twins we can put in there to total 8 are 1/7, 2/6 or 3/5.

We know that 5 and 6 have already been placed in that box, so the cage must contain 1/7.

Now look at the cage with a total of 6. The only twins which could be in there are 1/5 or 2/4, so it must be 2/4.

Now the cage of 17 must contain 8/9 so place those. So the empty cell must contain a 3.

We can also place twins of 2/4 in the cage of 6 in the bottom-left block — it can’t be 3/3, and we’ve already used the 5, so it can’t be 5/1.

Now we know a 3 exists in the cage of 14 which straddles the centre and the middleright boxes, we can use the technique of 45 and find the 3 in that centre box.

Equally, in the middle-left box the cage of 10 must contain 9 and 1 as it can only possibly be 1/9, 2/8, 3/7 or 4/6. 1, 3, 4 and 7 are already on that Sudoku line across and down, so it must be 9 with the 1 below.