2008 AMC 12A Problems/Problem 17

Problem

Let $a_1,a_2,\ldots$ be a sequence determined by the rule $a_n=a_{n-1}/2$ if $a_{n-1}$ is even and $a_n=3a_{n-1}+1$ if $a_{n-1}$ is odd. For how many positive integers $a_1 \le 2008$ is it true that $a_1$ is less than each of $a_2$, $a_3$, and $a_4$?

$\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004$

Solution

All positive integers can be expressed as $4n$, $4n+1$, $4n+2$, or $4n+3$, where $n$ is a nonnegative integer.

  • If $a_1=4n$, then $a_2=\frac{4n}{2}=2n<a_1$.
  • If $a_1=4n+1$, then $a_2=3(4n+1)+1=12n+4$, $a_3=\frac{12n+4}{2}=6n+2$, and $a_4=\frac{6n+2}{2}=3n+1<a_1$.
  • If $a_1=4n+2$, then $a_2=2n+1<a_1$.
  • If $a_1=4n+3$, then $a_2=3(4n+3)+1=12n+10$, $a_3=\frac{12n+10}{2}=6n+5$, and $a_4=3(6n+5)+1=18n+16$.

Since $12n+10, 6n+5, 18n+16 > 4n+3$, every positive integer $a_1=4n+3$ will satisfy $a_1<a_2,a_3,a_4$.

Since one fourth of the positive integers $a_1 \le 2008$ can be expressed as $4n+3$, where $n$ is a nonnegative integer, the answer is $\frac{1}{4}\cdot 2008 = 502 \Rightarrow D$.

Alternate Solution

After checking the first few $a_n$ such as $1$, $2$ through $7$, we can see that the only $a_1$ that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for $a_n=3$, we notice the sequence yields $10$, $5$, and $16$, a valid sequence.

So we can set up an equation, $3x + 1 = 2(2k - 1)$ where x is equal to $a_1$. Rearranging the equation yields $(3x + 3)/4 = k$. Experimenting yields that every 4th $x$ after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to $2008/4 = 502 \Rightarrow D$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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

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