2006 AMC 12B Problems/Problem 19
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 and in the form of , or .
We can conclude straight away that using the divisibility rule.
If , the number is not divisible by (unless it's , which is not divisible by ), which means there are no , , , or year olds on the car, but that can't be true, as that would mean there are less than kids on the car.
If , then the only possible number is . is divisible by , , and , but not by and , so that doesn't work.
If , then the only number is , also not divisible by or .
If , the only number is . It is divisible by , , , and .
Therefore, we conclude that the answer is
NOTE: Automatically, since there are 8 children and all of their ages are less than or equal to 9 and are different, the answer choices can be narrowed down to or .
Alternate Solution
We know that the number of the plate is divisible by and in the form , , or for distinct digits .
Using the divisibility rule for , we can conclude that .
We also know that the number of the plate is even, because you can only discard one number from the integers through , inclusive ( children, oldest is ), and there's always going to be an even number left.
If one of the children was years old, then the plate number would be divisible by both and . Thus, the units digit must be . Then, the possible form of the plate number would be , , or . The first case is not possible because is not a possible age for the father.
We have the remaining forms and .
We know that , so has to be either or . can't be because are distinct. So, . However, the number is not divisible by both and , so whichever number you discard, the other one will still be there. This creates a contradiction, so Case cannot be true.
Similarly to Case , we can determine that has to be . However, the number is not divisible by both and . Using the same logic in case , we conclude that Case cannot be true.
Since we have disproven all of our cases, we know that it is impossible for one of the children to be years old.
NOTE: This might look tedious, but it only took me around 30 seconds to do on paper.
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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