2006 AMC 10B Problem 25

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 $\boxed{\textbf{(B)}\ 5}$.

Divisibility by $5$ combined with evenness pins the last digit to $0$, and the repeated-digit structure then leaves almost no freedom, so the assumption "5 is present" collapses after checking one candidate number. In "which one is impossible" problems, attack the choice with the most restrictive divisibility rule first.

Constructive: the digit sum $2(a + b)$ must be a multiple of $9$, so $a + b = 9$. Among multiples of $504 = \mathrm{lcm}(1,2,3,4,6,7,8,9)$ with four digits, $5544$ is the only one with the two-digits-twice pattern, and it is not divisible by $5$. Checking lcms that omit $4$, $6$, $7$ or $8$ instead (each still needs $5 \mid n$, hence a $0$ digit) yields no valid plate.

• Testing digit-sum conditions only and never using that the number is even, or forgetting that exactly one age from 1 to 9 is absent.
• Concluding $4$ is missing because $9090$ fails divisibility by $4$, without noticing that then $8$ would have to divide the number.