# Smallest range

What is the smallest possible range that these 4 integers could have?

## Problem

Four integers are chosen so that their mean is 2017.

The integers are all different.

What is the smallest possible range that the 4 integers could have?

*This problem is taken from the UKMT Mathematical Challenges.*

## Student Solutions

**Balancing the sum of 4 numbers**

If the integers have a small range, then they are all close together. $2017$ is the mean, so they should all be close to $2017.$ If they were allowed to be the same, then they could all be $2017,$ with a range of $0,$ since:

$$(2017 + 2017 + 2017 + 2017)\div4 = 2017$$

We need to change the numbers in the sum so that they are all different to each other without changing the value of the sum. If we decrease one by $1$, then we can increase another by $1$ without affecting the value of the sum:

$$2017 + 2017 + 2017 + 2017=2016+2018+2017+2017$$

Now we still have two integers that are the same. This time we will have to decrease one of them by $2$ and increase the other by $2$:

$$2016+2018+2017+2017=2016+2018+2015+2019$$

So the smallest possible range is the difference between $2019$ and $2015$, which is $4$.

**Trying out consecutive integers**

To have the smallest range possible, the integers should be as close together as possible, so they should be consecutive.

$2017$ is the mean, so they should all be close to $2017$. So we could try sets of consecutive integers close to $2017$.

$(2014+2015+2016+2017)\div4=8062\div4=2015.5<2017$

$(2015+2016+2017+2018)\div4=8066\div4=2016.5<2017$

$(2016+2017+2018+2019)\div4=8070\div4=2017.5>2017$

So the integers cannot all be consecutive.

Notice that when $2017$ was in the second half of the four consecutive integers (so there were more numbers smaller than $2017$ than larger than $2017$), the mean was too small.

But when $2017$ was in the first half of the four consecutive integers (so there were more numbers larger than $2017$ than smaller than $2017$), the mean was too big.

By removing $2017$ from the four integers, we can choose a set which has two numbers smaller than $2017$ and two numbers larger than $2017$:

$(2015+2016+2018+2019)\div4=8068\div4=2017$.

So the smallest possible range is the difference between $2019$ and $2015$, which is $4$.

**Using algebra and factors**

To have the smallest range possible, the integers should be as close together as possible, so they should be consecutive. Four consecutive integers can be written as $n,n+1,n+2,n+3$.

So we want $$\begin{align}(n+(n+1)+(n+2)+(n+3))\div4&=2017\\ \Rightarrow n+n+1+n+2+n+3&=4\times2017\\ \Rightarrow4n+6&=4\times2017\end{align}$$

But $4n$ and $4\times2017$ are both multiples of $4$ (when $n$ is an integer), so the difference between them can't be $6$. So the four integers can't be consecutive. The difference between $4n$ and $4\times2017$ should be $4$ or $8,$ so we can tweak our integers to make this difference $4$ or $8.$

Suppose we choose to make the difference $8$, so our sum $n+(n+1)+(n+2)+(n+3)$ needs to increase by $2$. If we increase $n$ or $(n+1)$ by $2$, then we get $(n+2)$ or $(n+3)$ - but that's not allowed, because we've already used $(n+2)$ and $(n+3)$ and the integers are all supposed to be different. If we increase $(n+2)$ or $(n+3)$ by $2$, we get $(n+4)$ or $(n+5)$. To keep the range as small as possible, we should increase $(n+2)$ by $2$ to give $(n+4)$. So now we have: $$\begin{align}(n+(n+1)+(n+4)+(n+3))\div4&=2017\\

\Rightarrow n+n+1+n+4+n+3&=4\times2017\\

\Rightarrow 4n+8&=4\times2017\\

\Rightarrow 4n&=4\times2017-8\\

\Rightarrow 4n&=4\times2017-4\times2\\

\Rightarrow 4n&=4\times(2017-2)\\

\Rightarrow n&=2015\end{align}$$ So the integers are $2015,2016,2019,2018$ and so the range is $4$.