A student notices that in a list of five integers, the mean, median and mode were consecutive integers in ascending order. What is the largest range possible for these five integers and why? Please explain clearly.
When I first saw the question I thought that one of the integers (a mean, median, mode integer) had to be $x-1$, one had to be $x$ and one $x+1$. But I do not know which is which and how to find out the largest range possible
On
Call the five integers $r\le s\le t\le u\le v$. The mean is $x-1$; that gives us the equation, $$r+s+t+u+v=5x-5$$ The median is $x$; that tells us $t=x$. The mode is $x+1$; that tells us $u=v=x+1$. Putting these together, you get an equation relating $r,s,x$, and you get $s<x$. The range is $v-r$, which is $x+1-r$. Now putting these together, you can work out all the possible values of $r$, and then pick out the one that maximizes the range.
Can you fill in the details?
Building on Gerry's answer...
OK, I'll start from the beginning. Let's call the numbers $r,s,t,u$ and $v$ with $r\le s\le t\le u\le v$. The median is $x$, so $t=x$. The mode is $x+1$, so $u=x+1$. We need more than one with this value, so $v=x+1$ as well. The mean is $x-1$, so \begin{align}\frac{r+s+t+u+v}{5}&=x-1\\r+s+t+u+v&=5x-5\end{align}
The range is $v-r$ and we have already determined that $v=x+1$ so to maximize the range we need to minimize $r$. In order to do that we maximize $s$, so $s=x-1$. Now we just need to find $r$. Let $r=x-a$, then substitute into the above equation and solve for $a$: \begin{align}r+s+t+u+v&=5x-5\\(x-a)+(x-1)+x+(x+1)+(x+1)&=5x-5\\5x-a+1&=5x-5\\-a&=-6\\a&=6\end{align} So the maximum range is $v-r=(x+1)-(x-a)=1+a=1+6=7$