Bunuel
Let \(a_1\), \(a_2\), ... be a sequence determined by the rule \(a_n=\frac{a_{n-1}}{2}\) if \(a_{n-1}\) is even and \(a_n=3a_{n-1}+1\) if \(a_{n-1}\) is odd. For how many positive integers \(a_1 \leq 2008\) is it true that \(a_1\) is less than each of \(a_2\), \(a_3\), and \(a_4\)?
(A) 250
(B) 251
(C) 501
(D) 502
(E) 1004
We are given 2008 possible numbers, easiest way to filter out at the beginning is to do an odd even split and see if it helps us eliminate half the numbers because then we need to focus on only half remaining
Consider a1 as EVEN:
a1 = 2
a2 = 2/2 (Because a1 is even) = 1
Now we already have a1 is NOT LESS than a2 so we can stop there because we need to find numbers such that a1 is less than a2,a3 and a4
So we can safely assume that all options where a1 is even will not satisfy that because if a1 is even, then a2 = a1/2 which will be less than a1
So all a1 EVEN are out
We are left with 1004 options where a1 is odd. Let us test a few scenarios and see if all satisfy or part do!
a1 a2 a3 a4
1 4 2 1 NO
3 10 5 16 YES
5 16 8 4 NO
7 22 11 34 YES
So, we can see that every alternate term satisfies so we are left with half of the options so 502
Answer - D