2006 AMC 10B Problem 25

1. Which ages are there? Eight children with different ages, the oldest $9$. So the ages are eight of the nine numbers $1, 2, \ldots, 9$, and exactly one number is missing. The question asks which one.
2. The plate number is even. Among $2, 4, 6, 8$ at most one is missing, so some even age is present, and the plate number is a multiple of $2$.
3. Test the choice (B) first, because $5$ has the strictest rule: suppose $5$ is an age. A number that is a multiple of both $2$ and $5$ is a multiple of $10$, so the plate ends in $0$.
4. What pattern is left? The plate uses two digits, each twice. One of them is $0$; call the other $d$. The last digit is $0$, and the last two digits are Mr. Jones's age, so they are not $00$. That forces the pattern $d0d0$.
5. Use the oldest child, age $9$. A number is a multiple of $9$ exactly when its digit sum is. The digit sum of $d0d0$ is $2d$, and for a single digit $d$ that only works when $d = 9$. So the plate would be $9090$.
6. Check $4$ and $8$. For a multiple of $4$, the last two digits must form a multiple of $4$; $90$ is not. So $9090$ is not a multiple of $4$, and it is not a multiple of $8$ either. But at most one age is missing, so $4$ and $8$ cannot both be missing. This is impossible.
7. So $5$ is the missing age.

A plate that really works: the ages $1, 2, 3, 4, 6, 7, 8, 9$ all divide $504$, and $11 \cdot 504 = 5544$, which has two $5$s and two $4$s and ends in $44$.

The answer is $\boxed{\textbf{(B)}\ 5}$.

In a "which one is impossible" question, start with the choice that has the strictest rule. Being a multiple of both $2$ and $5$ fixes the last digit as $0$, and the "two digits, each twice" pattern then leaves only one possible number to check. Once that number fails, you are done.

Build the plate instead. Its digit sum is twice the sum of its two digits, and it must be a multiple of $9$, so the two digits add to $9$. If $5$ is missing, the plate must be a multiple of $504$ (the smallest number that $1, 2, 3, 4, 6, 7, 8, 9$ all divide). Going through the four-digit multiples of $504$, only $5544$ uses two digits twice each, and its digits $5 + 4 = 9$ fit. It is not a multiple of $5$, which matches (B).

The eight distinct ages are at most $9$ and include $9$, so they are $\{1, \ldots, 9\}$ with exactly one number removed. Call the plate number $n$.

Suppose $5$ is not the removed age. Only one of $2, 4, 6, 8$ can be missing, so $n$ is even; being also a multiple of $5$, it ends in $0$. Its two digits are therefore $0$ and some $d$, each used twice. The last two digits form Mr. Jones's age, so they are not $00$; the number is not $d00d$ either (it must end in $0$), leaving $n = d0d0$.

Divisibility by $9$ (the oldest child's age) forces $2d$ to be a multiple of $9$, so $d = 9$ and $n = 9090$. But $90$ is not a multiple of $4$, so $4 \nmid 9090$; then $4$ is the missing age, so $8$ must be present, yet $8 \nmid 9090$ either. Contradiction.

Hence $5$ is the missing age. (A consistent plate exists: the ages $1, 2, 3, 4, 6, 7, 8, 9$ have lcm $504$, and $11 \cdot 504 = 5544$ has two digits each twice and ends in $44$.)

The answer is $\boxed{\textbf{(B)}\ 5}$.