2006 AMC 10B Problems/Problem 25

Revision as of 12:42, 23 January 2012 by Minsoens (talk | contribs) (Solution)


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?

$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8$


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 $5 \not\in S$, 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 $5 \not\in S$ and $5$ is not the age of one of Mr. Jones' kids. $\boxed{ B }$

(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 number must be a multiple of $lcm(1,2,3,4,6,7,8,9) = 504$. Since $11\cdot504 = 5544$ 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 $5544$.)

Another easier way is that the possible ages for all the younger children are 8,7,6,5,4,3,2, and 1. Only one of these eight numbers isn't the age of one of the seven younger children. If one of the children is 5, then there must be a child who is either 8 or 4, because only one of these numbers cannot be an age. So if the license plate number is divisible by 5 and 4 or 8, it ends in 00 which isn't a possible age for the dad.

See also

2006 AMC 10B (ProblemsAnswer KeyResources)
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