Problem 637

Consider the following list of $5$ integers, ordered from smallest to
largest, and having a median of $10$:
$a,b,10,c,d$.

Since $a$ is the smallest integer in
the list and $d$ is the largest, and
the list has a range of $7$, then
$d$ is $7$ more than $a$.

Since $a$ and $d$ differ by $7$, then to find the smallest possible
value of $a$, we can find the
smallest possible value of $d$ and
subtract $7$.

The $5$ integers in the list are
different from one another, and so the smallest possible value of $c$ is 11 ($c$ must be greater than the median $10$), and the smallest possible value of
$d$ is thus $12$.

Since $d$ is $7$ more than $a$, then the smallest possible integer in
the list is $12-7=5$.

(We note that $5,b,10,11,12$, where
$b$ is greater than $5$ and less than $10$, is such a list.)