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 | First, The number of the plate is divisible by 9 and in the form of | ||
| − | aabb, abba or abab | + | aabb, abba or abab. |
| − | We can conclude straight away that a+b is 9 using the 9 divisibility rule | + | We can conclude straight away that a+b is 9 using the 9 divisibility rule. |
| − | + | If b=1, the number is not divisible by 2 (unless it's 1818, which is not divisible by 4), which means there are no 2, 4, 6, or 8 year olds on the car, but that can't be true, as that would mean there are less than 8 kids on the car. | |
| − | + | If b=2, then the only possible number is 7272. 7272 is divisible by 4, 6, and 8, but not by 5 and 7, so that doesn't work. | |
| − | If b= | + | If b=3, then the only number is 6336, also not divisible by 5 or 7. |
| − | If b= | + | If b=4, the only number is 5544. It is divisible by 4, 6, 7, and 8. |
| − | + | Therefore, we conclude that the answer is <math> \mathrm{(B)}\ 5 </math> | |
== 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 09:22, 17 February 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
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.
If b=1, the number is not divisible by 2 (unless it's 1818, which is not divisible by 4), which means there are no 2, 4, 6, or 8 year olds on the car, but that can't be true, as that would mean there are less than 8 kids on the car.
If b=2, then the only possible number is 7272. 7272 is divisible by 4, 6, and 8, but not by 5 and 7, so that doesn't work.
If b=3, then the only number is 6336, also not divisible by 5 or 7.
If b=4, the only number is 5544. It is divisible by 4, 6, 7, and 8.
Therefore, we conclude that the answer is 
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 | |
