2006 AMC 10B Problem 25

Problem 25 of 25HarderNumber Theory

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?

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Solution

Since a child is

9,

the number is divisible by

9,

so its digit sum

2(a+b)

is a multiple of

9,

which forces

a+b=9.

The eight distinct ages are eight of the nine integers from

1

through

9,

so at least one of

4

and

8

is an age. Hence the plate number is divisible by

4.

Up to interchanging the two digits, its pattern is

aabb,

abab,

baab.

Combining these patterns with

a+b=9

and divisibility by

4

leaves

\begin{gathered} 1188,\ 2772,\ 3636,\ 5544,\\ 6336,\ 7272,\ 9900. \end{gathered}

The last candidate would make Mr. Jones’s age

00,

so it is impossible. None of the other six candidates is divisible by

5,

5

cannot be one of the children’s ages. The conditions are attainable:

5544

is divisible by each age in

\{1,2,3,4,6,7,8,9\}

and its last two digits give Mr. Jones’s age as

44.

Thus, the correct answer is B.

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