$\begin{matrix} 32 \times 38 & = & 30 \times 40 + 2 \times 8 \\ 34 \times 36 & = & 30 \times 40 + 4 \times 6 \\ 56 \times 54 & = & 50 \times 60 + 6 \times 4 \\ 73 \times 77 & = & 70 \times 80 + 3 \times 7 \end{matrix}$

Chloe Pattenden and Nina Williams (Y7) St James Middle School, Bury St Edmunds found these examples:

$\begin{matrix} 81 \times 89 & = & 80 \times 90 + 1 \times 9 = 7209 \\ 72 \times 78 & = & 70 \times 80 + 2 \times 8 = 5616 . \end{matrix}$

HOW WE WORKED IT OUT: The units on the first two numbers have to add up to 10 to make this work and the second pair of numbers on the right hand side also have to add up to 10. The tens in the first product on the right hand side have to be 10 apart.

Suzanne Abbott & Nisha Doshi, (Y10) The Mount School, York proved that the method always works. Here is their solution:

Yes these are all correct. Here are some more examples

$\begin{matrix} 51 \times 59 & = & 50 \times 60 + 1 \times 9 \\ 45 \times 45 & = & 40 \times 50 + 5 \times 5 \end{matrix}$

It always seems to work. Using algebra, we can try to generalise this

$\begin{matrix} (10 a + b) (10 a + c) & = & 10 a \times 10 (a + 1) + bc \\ 100 a^{2} + 10 ac + 10 ab + bc & = & 100 a^{2} + 100 a + bc \\ 10 ac + 10 ab & = & 100 a \\ 10 a (b + c) & = & 100 a \\ so b + c & = & 10 \end{matrix}$

This fits with the examples we were given and the ones we have chosen.

This proof depends on the argument being reversible. Set out slightly differently it shows that the method works if and only if $b + c = 10$ and it is perfectly rigorous.

$\begin{matrix} (10 a + b) (10 a + c) & = & 100 a^{2} + 10 ab + 10 ac + bc \\ = & 100 a^{2} + 100 a + bc + 10 ab + 10 ac - 100 a \\ = & 100 a^{2} + 100 a + bc + 10 a (b + c - 10) \\ = & 10 a \times 10 (a + 1) + bc \end{matrix}$