Skip to main content
### Number and algebra

### Geometry and measure

### Probability and statistics

### Working mathematically

### For younger learners

### Advanced mathematics

# What Does it All Add up To?

## You may also like

### Adding All Nine

### Old Nuts

Links to the University of Cambridge website
Links to the NRICH website Home page

Nurturing young mathematicians: teacher webinars

30 April (Primary), 1 May (Secondary)

30 April (Primary), 1 May (Secondary)

Or search by topic

Age 11 to 18

Challenge Level

- Problem
- Student Solutions
- Teachers' Resources

Tobias from Wilson's School sent in this diagrammatic proof that the sum of four consecutive numbers can never be a multiple of 4:

Shayen from Wilson's School in England investigated the sums of 5 consecutive numbers. Here is Shayen's work (click to see a larger version):

Tobias sent in this diagrammatic proof that what Shayen noticed is always true:

Sunhari from British School Muscat and Mahdi from Mahatma Gandhi International School in India sent in this algebraic proof:

Let the five consecutive numbers be $n, n+1, n+2, n+3, n+4$

$\begin{split}n+n+1+n+2+n+3+n+4 & =5n+10\\

&=5(n+2)\\

&=5a for a = n+2\end{split}$

Shayen also investigated the sums of 6 consecutive numbers. Here is Shayen's work (click to see a larger version):

Here is Tobias's diagrammatic proof of what Shayen noticed:

Here is Mahdi and Sunhari's algebraic proof:

Let the six consecutive numbers be $n, n+1, n+2, n+3, n+4, n+5$

$\begin{split}n+n+1+n+2+n+3+n+4+n+5&= 6n + 15\\&= 3(2n+5)\end{split}$

It is an odd multiple of 3, and cannot be factorised to a multiple of six.

Shayen made a conjecture about adding other numbers of consecutive numbers (click to see a larger version):

Egeman from Wilson's School represented this algebraically:

Let $a$ be the first number in a sequence of consecutive numbers

Let $n$ be the number of numbers in a sequence of consecutive numbers

Let $s$ be the sum of the numbers in a sequence of consecutive numbers

where $n=1, s=1a+0$;

where $n=2, s=2a+1$;

where $n=3, s=3a+3$;

where $n=4, s=4a+6$;

where $n=5, s=5a+10$;

where $n=6, s=6a+15$; etc

$s$ consists of two parts: a multiple of $a$ and a triangle number.

The multiple of $a$ is always $na$.

The triangle number is the $(n-1)^\text{th}$ term in the sequence of triangle numbers.

Mahdi and Sunhari showed why $s$ takes this form. This is Mahdi's work:

From here, Mahdi, Sunhari and Egeman all used a formula for the $(n-1)^\text{th}$ triangular number to show that if $n$ is odd, $s$ is a multiple of $n$, and if $n$ is even, then $s$ is not a multiple of $n$ but is a multiple of $\frac n2.$ Here is Mahdi's work:

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!

In turn 4 people throw away three nuts from a pile and hide a quarter of the remainder finally leaving a multiple of 4 nuts. How many nuts were at the start?