imc

Let $S$ be the set of the ages of Mr. Jones' children (in other words $i\in S$ if Mr. Jones has a child who is $i$ years old). Then $|S|=8$ and $9\in S$. Let $m$ be the positive integer seen on the license plate. Since at least one of $4$ or $8$ is contained in $S$, we have $4|m$. We would like to prove that $5\notin S$, so for the sake of contradiction, let us suppose that $5\in S$. Then $5\cdot 4=20|m$ so the units digit of $m$ is $0$. Since the number has two distinct digits, each appearing twice, another digit of $m$ must be $0$. Since Mr. Jones can't be $00$ years old, the last two digits can't be $00$. Therefore $m$ must be of the form $d0d0$, where $d$ is a digit. Since $m$ is divisible by $9$, the sum of the digits of $m$ must be divisible by $9$ (see Divisibility rules for 9). Hence $9|2d$ which implies $d=9$. But $m=9090$ is not divisible by $4$, contradiction. So $5\notin S$ and $5$ is not the age of one of Mr. Jones' kids. $\boxed{5}$ (We might like to check that there does, indeed, exist such a positive integer $m$. If $5$ is not an age of one of Mr. Jones' kids, then the license plate's number must be a multiple of $lcm(1,2,3,4,6,7,8,9)=504$. Since $11\cdot 504=5544$ and $5544$ is the only $4$ digit multiple of $504$ that fits all the conditions of the license plate's number, the license plate's number is $5544$.)