Difference between revisions of "2006 AMC 10B Problems/Problem 25"
(→See also) |
|||
Line 16: | Line 16: | ||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 12:18, 4 July 2013
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?
Solution
Let be the set of the ages of Mr. Jones' children (in other words if Mr. Jones has a child who is years old). Then and . Let be the positive integer seen on the license plate. Since at least one of or is contained in , we have .
We would like to prove that , so for the sake of contradiction, let us suppose that . Then so the units digit of is . Since the number has two distinct digits, each appearing twice, another digit of must be . Since Mr. Jones can't be years old, the last two digits can't be . Therefore must be of the form , where is a digit. Since is divisible by , the sum of the digits of must be divisible by (see Divisibility rules for 9). Hence which implies . But is not divisible by , contradiction. So and is not the age of one of Mr. Jones' kids.
(We might like to check that there does, indeed, exist such a positive integer . If is not an age of one of Mr. Jones' kids, then the license plate number must be a multiple of . Since and is the only digit multiple of that fits all the conditions of the license plate's number, the license plate's number is .)
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 10 Problems and Solutions |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.