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2006 AMC 10B Problems/Problem 25

Revision as of 00:17, 23 February 2011 by Fibonacci97 (talk | contribs) (→‎Solution: Simpler solution for problem 25)

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

Since all the children have different ages, and the oldest child is $9$, only $1$ of the numbers between $1$ and $8$ inclusive cannot be the age of a child.

To be evenly divisible by the age of all the kids, the license plate number must be a multiple of the least common multiple of the ages of all the kids.

Since being divisible by $5$ restricts the possibilities of the digits in the number, let's analyze the possibilities of license plate numbers if one of Mr. Jones's kids is $5$ years old.

Since the age of one of Mr. Jones's kids must be an even number, the number must be divisible by $10$.

Since the number is divisible by $10$, the units digit must be $0$

Since the number has two distinct digits, each appearing twice, another digit must be $0$

Since Mr. Jones can't be $00$ years old, the last two digits can't be $00$

Therefore, the number must be in the form $d0d0$, where $d$ is a digit.

Since the oldest child is $9$, the number must be divisible by $9$.

By divisibility rules, the sum of the digits must be a multiple of $9$

$2d \equiv 0\bmod{9}$

$d = 0$ or $d = 9$

Since the number has two distinct digits, $d$ can't be $0$.

So $d = 9$, and the only possible number is $9090$.

Since either $4$ or $8$ is one of the kids' ages, the number must be divisible by $4$.

Since $9090$ is not divisible by $4$, it is not a possible number.

Therefore, there are no possible license plate numbers if $5$ is an age of one of Mr. Jones's kids. So $5$ can't be the age of one of Mr. Jones's kids.

If $5$ is not an age of one of Mr. Jones's 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 $5544$ and the number that is not an age of one of Mr. Jones's kids is $5 \Rightarrow B$


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

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