2006 AMC 12B Problem 19

Below is the professionally curated solution for Problem 19 of the 2006 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

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Concepts:divisibilitydigitscasework

Difficulty rating: 1920

19.

Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9,9, spots a license plate with a 44-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?

44

55

66

77

88

Solution:

The number has the form aabb,aabb, abab,abab, or baab.baab. Divisibility by 99 means 2(a+b)2(a + b) is a multiple of 9,9, so a+b=9.a + b = 9.

The children include a 44- or 88-year-old, so the number is divisible by 4.4. The possibilities become 1188,2772,3636,5544,6336,7272,9900.1188, 2772, 3636, 5544, 6336, 7272, 9900.

Since the last two digits are Mr. Jones's age, 99009900 is impossible, and none of the others is a multiple of 5.5. So the children's ages cannot include 5.5. Indeed 55445544 is divisible by 1,2,3,4,6,7,8,9.1, 2, 3, 4, 6, 7, 8, 9.

Thus, the correct answer is B.

Problem 19 in Other Years