Difference between revisions of "2006 AMC 12B Problems/Problem 19"

(Solution 2)
(Solution 1)
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== Solution ==
 
== Solution ==
===Solution 1===
 
 
First, The number of the plate is divisible by 9 and in the form of
 
First, The number of the plate is divisible by 9 and in the form of
 
aabb, abba or abab
 
aabb, abba or abab

Revision as of 19:21, 15 January 2012

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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

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 (ProblemsAnswer KeyResources)
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