Summing Consecutive Numbers

Liam noticed that:

If you add an odd number to an even number it is always odd - so you can't make 2 consecutive numbers add up to an even number.
 

Several of you, including Luke, Nikhil, Hari, Carthick, William, Daniel, Babu and Alex, noticed: 

You can make multiples of $3$ using $3$ consecutive numbers.

Sonia from the Garden International School added:

I've found out that since 1+2=3
2+3 must add up to 2 more.

So if 2 consecutive numbers are added together, the totals will go up by 2.

That means if 3 numbers are added together (1+2+3), the next numbers (2+3+4) will add up to 3 more. So if 3 consecutive numbers are added together, the totals will go up by 3.

It means there is a pattern like 'if 4 numbers are added together, the totals will go up by 4'.

Owen and Thomas said:

All except the prime number $2$. Can you think why?

Year 7 students at Streetfield Middle School and students from Mount Pleasant Primary School noticed that:

Any number which is a power of $2$ cannot be made with consecutive sums:

$2, 4, 8, 16, 32 , 64, 128...$

Rajeev from Haberdashers' Aske's Boys' School sent us this clear and comprehensive solution.

Isabelle, Chip and Sam all had similar ideas about summing an odd number of consecutive integers. Chip explained his reasoning clearly: 

The consectutive integers can be found by dividing any number by one of its odd factors.

For example, dividing $33$ by $3$ gives us $11$.
Then, add $1$ and take $1$ on either side until there are as many numbers as the factor you used.
So $10$, $11$ and $12$ are the three numbers from dividing $33$ by $3$.

If you then say $10$ is $1+2+3+4$ but it is not divisible by $4$ that only works because it is actually being divided by $5$ but the zero is left out.

This method works to any size number for example $91 = 13 \times7$ so it can be made from $10 + 11 + 12 + 13 + 14 + 15 + 16$.

Even numbers only divide odd numbers and work correctly as $10$ divided by $2$ equals $5$ which goes to $6 + 4 = 10$ but they are not consecutive.

If therefore you divide $39$, an odd, by $6$, an even, you get $6.5$. Then using some common sense you get $4$, $5$, $6$, $7$, $8$, $9$.

The only way to tell if the factor will work is to see if the intended numerator is halfway between multiples of the factor.
For example, with $39$ it is halfway between $36$ and $42$, both multiples of $6$ so it will be able to have six consecutive numbers add up to it.

For those of you who want to investigate this problem further the following algebraic solution from Jack, Paul and Matthew might be helpful:

Like the number $15$, all odd numbers can be written as the sum of two consecutive integers.
For any odd number $k$, we know that $k-1$ and $k+1$ are even,
also $(k-1)/2$ and$(k+1)/2$ are consecutive integers,
so we can write $k=(k-1)/2+(k+1)/2$.

All numbers of the form $N = pk$ where $p$ and $k$ are integers, and $k$ is odd and greater than one, can be written as the sum of consecutive integers.
As before we use the integers $(k -1)/2$ and $(k +1)/2$ but this time we have to add $2p$ consecutive integers, $p$ of them less than $(k /2)$ and $p$ of them greater than $(k /2)$, so that their mean is $k /2$ and their sum is $(2 p ) \times( k /2) = pk$
For example $N = 44 = 4 \times 11$ can be written as the sum of $8$ consecutive integers, $4$ less than $5.5$ and $4$ greater than $5.5$ so that the sum is $8 \times5.5 = 44$.
 

$44$ = ${\bf \frac{11 - 7}{2}}$+ ${\bf \frac{11 - 5}{2}}$ + ${\bf \frac{11 - 3}{2}}$ + ${\bf \frac{11 - 1}{2}}$ + ${\bf \frac{11 + 1}{2}}$ + ${\bf \frac{11 + 3}{2}}$ + ${\bf \frac{11 + 5}{2}}$+ ${\bf \frac{11 + 7}{2}}$

$= 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 $

This construction works for all numbers of the form $N = pk$ where $p$ and $k$ are integers and $k$ is odd and greater than one, but the consecutive sum may include some negative integers. If this happens then a string of terms reduces to zero giving fewer terms in the final expression. By this method, \begin{eqnarray}12 &=& 4 \times3 \\ &=& (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 \\ &=& 3 + 4 + 5 \end{eqnarray}
giving three terms in the final expression rather than eight.


If a number is a power of $2$ then it is impossible to write it as a sum of consecutive integers:

Consider the sum of consecutive integers starting at $( q + 1)$ and ending at $p$ . There are $( p - q )$ terms in this sum. The mean of the terms is $( p + q + 1)/2$ so we can write: $$( q + 1) + ( q + 2) + \ldots+( p - 1) + p = ( p - q )( p + q + 1)/2$$ As $( p - q )$ and $( p + q + 1)$ cannot both be even, this means that the sum of consecutive integers must represent a number which has at least one odd factor (other than the factor $1$) so a power of $2$ cannot be written as a sum of consecutive integers.
We use the fact here that if $(p - q)$ is odd then $(p + q)$ is odd and $(p + q + 1)$ is even, or if $(p - q)$ is even then $(p + q)$ is even and $(p + q + 1)$ is odd.