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Well done to Agata and Elliot from England, Takuto from Bangkok Patana School in Thailand, Kabir, Arnav P., Arnav B., Nithya, Deethya, Mithravinda, Rivaan, Uday, Udit, Akshobhyan, Srishti, Ruhi, Tejas, Valerie, Nandiinii and Padmini from Ganit Kreeda, Vicharvatika in India and Mason from Irymple Secondary College in Australia who all sent in correct or nearly correct completed grids. This is Takuto's grid, which is fully correct:
Mason described using different approaches, and finding better methods:
First I started guessing with the factors which now I realise is inefficient and time-consuming if it wasn’t for the help of Julie Powell I would have continued and it would have potentially taken me longer to finish this problem. Julie taught me the importance of factor trees, which sped up the process much faster. I also had times when there were multiple options so I would revert to the definite ones. I then considered squares that overlapped as they limited potential options, it was in them that I began to find my solution. The smaller prime numbers such as 3 were really helpful because they only had one option of 1 $\times$ 1 $\times$ 3. I also realised that the products that had 5 in them would have to have a factor of 5 otherwise it would be impossible to make that product.
Once again, I would like to thank Julie Powell for the few hints and suggestions along the way.
Here are some of Mason's factor trees:
Shubhangee covered the Ganit Kreeda students' journey in tabular form:
Recording of moves is also done. But, it’s not complete. The entries shown in pitch colour are revisited back. This is the overall flow, where I tried to explain each and every move logically.
Now we can solve remaining sudoku using all rules of normal sudoku.
C8 & H9 = 5 (Completed columnwise)
I4 = 1 (Columnwise) & G7 = 1 (Boxwise)
F4=3, B5=3 and A8 = 3 (Tried to put 3 columnwise and boxwise)
After this we were stuck for longer time. But then instead of writing all possible numbers in each cell, we changed our approach. We just wrote wherever we can write 4 for each box. Then we checked for each column and it was clear that for 7th column there is only one place for 4 which is C7.
Then it was pretty smooth to complete the puzzle.
Elliot's journey to complete the product sudoku was different. This is Elliot's work:
This is what my grid looked like when I started doing normal sudoku at the end of the steps that I wrote out:
This is what it looked like when I got stuck, when I realised that there was nowhere in the bottom left sub-grid that 3 could go (I used grey for possibilities and black for definites):
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.