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Let the median be $n$, so the numbers are:
The mode is $n-1$, so the smallest two numbers must both be $n-1$, giving us:
The mean is $n+1$, so the total of the five numbers is $5n+5$, so the last two numbers must add up to $2n+7$.
The fourth number must be greater than $n$ (if it was $n$ there would not be a unique mode) so it must be at least $n+1$. That would give a value of $n+6$ for the fifth number.
If the fourth number was any bigger, the fifth number would have to be smaller (giving us a smaller range), so this gives the maximum range:
The difference between $n+6$ and $n-1$ is $7$,
so the maximum range is 7.
Possible Range
Age 14 to 16
ShortChallenge Level 
Let the median be $n$, so the numbers are:
_ , _ , $n$, _ , _
The mode is $n-1$, so the smallest two numbers must both be $n-1$, giving us:
$n-1, n-1, n,$ _ , _
The mean is $n+1$, so the total of the five numbers is $5n+5$, so the last two numbers must add up to $2n+7$.
The fourth number must be greater than $n$ (if it was $n$ there would not be a unique mode) so it must be at least $n+1$. That would give a value of $n+6$ for the fifth number.
If the fourth number was any bigger, the fifth number would have to be smaller (giving us a smaller range), so this gives the maximum range:
$n-1, n-1, n, n+1, n+6$
The difference between $n+6$ and $n-1$ is $7$,
so the maximum range is 7.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
