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Double Digit

Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?


Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

Big Powers

Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.

Quick Times

Age 11 to 14
Challenge Level

$$\begin{eqnarray} 32\times 38 &=& 30\times 40 + 2\times 8 \\ 34\times36 &=& 30\times40 + 4\times6\\ 56\times54 &=& 50\times60 + 6\times4\\ 73\times77 &=& 70\times80 + 3\times7\\ \end{eqnarray}$$

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

$$\begin{eqnarray} 81\times89 &=& 80\times90 + 1\times9 = 7209 \\ 72\times78 &=& 70\times80 + 2\times8 = 5616.\\ \end{eqnarray}$$

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 and 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

$$\begin{eqnarray} 51\times59 &=& 50\times60 + 1\times9 \\ 45\times45 &=& 40\times50 + 5\times5\\ \end{eqnarray}$$

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

$$\begin{eqnarray} (10a + b)(10a + c) &=& 10a \times10(a + 1) + bc \\ 100a^2 +10ac +10ab + bc &=& 100a^2 + 100a + bc\\ 10ac + 10ab &=& 100a\\ 10a(b + c) &=& 100a\\ so b + c &=& 10\\ \end{eqnarray}$$

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.

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