2006 AMC 12B Problems/Problem 19

Problem

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?

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

Solution

First, The number of the plate is divisible by $9$ and in the form of $aabb$ , $abba$ or $abab$ .

We can conclude straight away that $a+b= 9$ using the $9$ divisibility rule.

If $b=1$ , the number is not divisible by $2$ (unless it's $1818$ , which is not divisible by $4$ ), which means there are no $2$ , $4$ , $6$ , or $8$ year olds on the car, but that can't be true, as that would mean there are less than $8$ kids on the car.

If $b=2$ , then the only possible number is $7272$ . $7272$ is divisible by $4$ , $6$ , and $8$ , but not by $5$ and $7$ , so that doesn't work.

If $b=3$ , then the only number is $6336$ , also not divisible by $5$ or $7$ .

If $b=4$ , the only number is $5544$ . It is divisible by $4$ , $6$ , $7$ , and $8$ .

Therefore, we conclude that the answer is $\mathrm{(B)}\ 5$

NOTE: Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to $5$ or $8$ .

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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2006 AMC 12B Problems/Problem 19

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