2008 AMC 12A Problems/Problem 17
Contents
Problem
Let be a sequence determined by the rule if is even and if is odd. For how many positive integers is it true that is less than each of , , and ?
Solution 1
All positive integers can be expressed as , , , or , where is a nonnegative integer.
- If , then .
- If , then , , and .
- If , then .
- If , then , , and .
Since , every positive integer will satisfy .
Since one fourth of the positive integers can be expressed as , where is a nonnegative integer, the answer is .
Solution 2
After checking the first few such as , through , we can see that the only that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for , we notice the sequence yields , , and , a valid sequence.
So we can set up an equation, where x is equal to . Rearranging the equation yields . Experimenting yields that every 4th after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to .
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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