### Summing Consecutive Numbers

Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?

### Always the Same

Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?

### Fibs

The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?

# Quick Times

##### Stage: 3 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.

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