AMC 10 · 2006 · #25
Grade 6 number-theoryMr. 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?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Mr. Jones has eight children, all different ages, and the oldest is $9$. A $4$-digit license plate uses exactly two different digits, each appearing twice (for example a number shaped like $aabb$ or $abab$). This number is divisible by every one of the eight children's ages, and its last two digits equal Mr. Jones's own age. Among the choices $4,5,6,7,8$, decide which one is $\textbf{not}$ the age of any of his children.
Givens: There are $8$ children, all of different ages, and the oldest is $9$ years old; The plate is a $4$-digit number using exactly two distinct digits, each appearing exactly twice; The plate is divisible by the age of every child; The last two digits of the plate equal Mr. Jones's age; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: Which of $4,5,6,7,8$ is not one of the eight children's ages
Understand
Restated: Mr. Jones has eight children, all different ages, and the oldest is $9$. A $4$-digit license plate uses exactly two different digits, each appearing twice (for example a number shaped like $aabb$ or $abab$). This number is divisible by every one of the eight children's ages, and its last two digits equal Mr. Jones's own age. Among the choices $4,5,6,7,8$, decide which one is $\textbf{not}$ the age of any of his children.
Givens: There are $8$ children, all of different ages, and the oldest is $9$ years old; The plate is a $4$-digit number using exactly two distinct digits, each appearing exactly twice; The plate is divisible by the age of every child; The last two digits of the plate equal Mr. Jones's age; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #4 Introduce a Variable, #14 Extreme Principle, #5 Look for a Pattern, #6 Guess and Check
The question is 'which one is NOT an age,' so Tool #3 (Eliminate Possibilities) fits: show one candidate is impossible and it must be the answer. Tool #4 (Introduce a Variable) names the two repeated digits $a$ and $b$, which turns the divisibility-by-$9$ rule into a clean equation about $a+b$. Tool #14 (Extreme Principle) says: don't test the candidates blindly — attack the one with the harshest divisibility rule first. A multiple of $5$ must end in $0$ or $5$, the strictest last-digit condition of all the choices, so $5$ is the case most likely to break. Chasing that break is exactly what cracks the problem.
Execute — Answer: B
4.OA.B.4 Step 1 Fix the age set and the digit shape
- Eight different ages, none above $9$, must be the numbers $1$ through $9$ with exactly one number missing.
- The oldest child is $9$, so $9$ is always an age; the missing number is one of $1$ through $8$.
- The plate is divisible by every age, so it is a common multiple of all eight ages.
- Write the plate's two repeated digits as $a$ and $b$ (each used twice), so the sum of its four digits is $a+a+b+b=2(a+b)$.
💡 Eight distinct ages capped at nine leave room to drop only one value, and being divisible by every age means being a multiple of all of them at once.
4.OA.B.4 Step 2 Divisibility by 9 forces a + b = 9
- Because $9$ is always a child's age, the plate is divisible by $9$, so its digit sum $2(a+b)$ is a multiple of $9$.
- Since $2$ shares no factor with $9$, the factor of $9$ must live in $a+b$, so $a+b$ is a multiple of $9$.
- The two digits are different, so the largest $a+b$ can be is $9+8=17$; the only multiple of $9$ it can equal is $9$.
- Thus $a+b=9$.
💡 A number is a multiple of nine exactly when its digits add to a multiple of nine, and two different digits can only reach nine, not eighteen.
4.OA.B.4 Step 3 Test the age 5: it forces the digits to be 0 and 9
- Suppose $5$ is one of the ages.
- Then the plate is divisible by $5$, so it ends in $0$ or $5$.
- Only one age is missing, so among the even ages $2,4,6,8$ at least three survive — the plate is even and cannot end in $5$.
- So it ends in $0$.
- The last digit $0$ must be one of the two repeated digits, so $0$ is used twice.
- Its partner $b$ then satisfies $0+b=9$, giving $b=9$.
- So the four digits would have to be $0,0,9,9$.
💡 Divisible by five and by two means the number must end in zero, and the digit-sum rule then pins the only possible partner digit.
4.NBT.B.6 Step 4 No plate of 0,0,9,9 works, so 5 is impossible
- An even $4$-digit number made of $0,0,9,9$ that ends in $0$ is either $9900$ or $9090$.
- But $9900$ ends in $00$, which would make Mr.
- Jones $0$ years old — impossible.
- And the plate must be divisible by $4$: whichever of $4$ or $8$ is present forces $4\mid N$ (and only one age is missing, so at least one of them is present), yet $9090$ is not divisible by $4$ since $90$ is not.
- Both options fail, so $5$ cannot be an age.
- By elimination, the age that is not one of the children's is $5$, which is choice (B).
💡 Every candidate plate built from those digits either ages the father to zero or fails the divisible-by-four test, leaving no legal number.
6.NS.B.4 Step 5 Confirm 5 is the missing age with a real plate
- If $5$ is the missing age, the ages are $1,2,3,4,6,7,8,9$.
- Their least common multiple is $\operatorname{lcm}=2^3\cdot 3^2\cdot 7=504$, so the plate must be a multiple of $504$.
- The four-digit multiple $504\times 11=5544$ uses exactly two digits, $5$ and $4$, each twice, and its last two digits $44$ make a believable age for Mr.
- Jones.
- So such a plate really exists, confirming that leaving out $5$ is consistent while leaving out any other choice is not.
💡 Building an actual working plate for the leave-out-five case proves the impossibility argument wasn't a fluke.
4.OA.B.4 Eight different ages, none above $9$, must be the numbers $1$ through $9$ with e 4.OA.B.4 Because $9$ is always a child's age, the plate is divisible by $9$, so its digit 4.OA.B.4 Suppose $5$ is one of the ages. Then the plate is divisible by $5$, so it ends i 4.NBT.B.6 An even $4$-digit number made of $0,0,9,9$ that ends in $0$ is either $9900$ or 6.NS.B.4 If $5$ is the missing age, the ages are $1,2,3,4,6,7,8,9$. Their least common mu Review
Reasonableness: The logic hangs on two unbreakable rules. Divisibility by $9$ (guaranteed, since a $9$-year-old exists) forces the two digits to add to $9$. Divisibility by $5$ would force the digits to be $0$ and $9$, and no arrangement of $0,0,9,9$ can be a real plate — it either makes the father $0$ years old or fails divisibility by $4$. Meanwhile the concrete plate $5544$ shows the leave-out-$5$ world is perfectly consistent: $5544=504\times 11$ is divisible by $1,2,3,4,6,7,8,9$ but not by $5$. Everything points to $5$.
Alternative: Skip straight to building the number. The other choices $4,6,8$ are covered automatically once $8$ and $9$ divide the plate ($8$ brings a $4$, and $2\cdot3$ brings a $6$), and $7$ and $9$ are forced too, so the real tension is only $5$ versus the rest. Asking for the smallest four-digit multiple of $\operatorname{lcm}(1,2,3,4,6,7,8,9)=504$ that repeats two digits lands on $5544$ directly, and since $5544$ is not divisible by $5$, the odd-one-out is $5$.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Treating the plate as a common multiple of all eight ages and using the digit-sum test for multiples of 9 to force $a+b=9$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Checking divisibility of the candidate plates 9900 and 9090 by 4 to rule out every arrangement of the digits 0,0,9,9.)6.NS.B.4Find greatest common factor and least common multiple of two numbers (Computing $\operatorname{lcm}(1,2,3,4,6,7,8,9)=504$ and finding the multiple $5544$ that confirms 5 is the missing age.)
⭐ A $9$-year-old forces the two repeated digits to add to $9$, and a $5$ would force the digits to be $0$ and $9$ — which can never make a real plate — so $5$ is the age nobody has, and $5544$ proves it.
⭐ A $9$-year-old forces the two repeated digits to add to $9$, and a $5$ would force the digits to be $0$ and $9$ — which can never make a real plate — so $5$ is the age nobody has, and $5544$ proves it.
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