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Thank you to Isabelle from Maidstone Grammar School for Girls who sent a correct solution for the completed square. Andrew, Brian, Dylan and Maroun from Greenacre Public School in Australia and Lawrence from the UK sent in very similar solutions that were clearly explained. This is Lawrence's solution:

1.  We started off by thinking about what digit could go in the top LH corner (A,N) and soon realised that it had to be 1 otherwise the multiple of 6 would not fit in the table.

2.  We then looked at the RH column and realised that as 5N would have to end in 5 or 0, the only place a 1 could go in the RH column was the only other odd multiple left, which was 3 (F,3).
The only number to give a multiple ending in 1 is 3 x 7, so (F,N) was 7, and we were then able to fill in the rest of column F and carry the 10s over into column E.

  A B C D E F
N 1         7
2N           4
3N           1
4N           8
5N           5
6N           2

3.  Now that we had all the digits we went back to column A and put the digits in, in numerical order (checking that it was feasible), which allowed us to work out that (B, N) was 4.

  A B C D E F
N 1 4       7
2N 2         4
3N 4         1
4N 5         8
5N 7         5
6N 8         2

4.  We then used trial and error to put the remaining 3 digits of N in starting at the RHS of the table and working down each column.

  A B C D E F
N 1 4 2 8 5 7
2N 2 8 5 7 1 4
3N 4 2 8 5 7 1
4N 5 7 1 4 2 8
5N 7 1 4 2 8 5
6N 8 5 7 1 4 2

We only realised when we had finished that there is a pattern in the rows, i.e. the digits are always in the same order: 1, 4, 2, 8, 5, 7.