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Consider the prime factorisation of 12:
$$\begin{align} 12 &= 2 \times 6 \\ &= 2 \times 2 \times 3 \end{align}$$
So $12 = 2^2 \times 3^1$.
Can you see how the table below can be used to find the six factors of 12?
$2^0$ 


$2^1$ 


$2^2$ 

The first branch gives us $2^0 \times 3^0 =1$
The second branch gives us $2^0 \times 3^1 =3$
The third branch gives us $2^1 \times 3^0 =2$
The fourth branch gives us $2^1 \times 3^1 =6$
The fifth branch gives us $2^2 \times 3^0 =4$
The sixth branch gives us $2^2 \times 3^1 =12$
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!
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?