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2008 AMC 12A Problems/Problem 17

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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 $\frac{1}{4}\cdot 2008 = 502 \Rightarrow D$ .

See Also

2008 AMC 12A (Problems • Resources)
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

Collatz Problem

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