Problem 805

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.)