2006 AMC 10B Problem 25

P21-25~7 minPrint

Divisibility & Factors Bases & Digits GCD & LCM Use the answer choices (mod checks, size, form) to eliminate or select Organized listing / direct enumeration

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children?

Show solution without answering

Key insight

If 5 were an age, the even plate number would end in 0 and look like d0d0. The rule for 9 forces 9090, which 4 does not divide.

Solution

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.
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 $2$ .
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 $0$ .
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 $d0d0$ .
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 $9090$ .
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.
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 $\boxed{\textbf{(B)}\ 5}$ .

Why this works

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.

Alternative approach

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).

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 8: Base Numbers
Introduction to Number Theory, Chapter 3: Multiples and Divisors (§3.3, §3.4, §3.7)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)