Difference between revisions of "2006 AMC 12B Problems/Problem 19"
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== Solution == | == 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 is 9 using the 9 divisibility rule | ||
| + | After trying various numbers, and reaching five | ||
| + | I can observe that b=0 or 5. | ||
| + | If b=5, the number is not divisible by 2 | ||
| + | If b=0, the number must be 9900, and is not divisible by 8 | ||
| + | Answer B 5 | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}} | ||
Revision as of 20:14, 1 February 2010
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Problem
Mr. Jones has eight children of different ages. On a family trip his olderst 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?
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 is 9 using the 9 divisibility rule After trying various numbers, and reaching five I can observe that b=0 or 5. If b=5, the number is not divisible by 2 If b=0, the number must be 9900, and is not divisible by 8 Answer B 5
See also
| 2006 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
