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Difference between revisions of "2008 AMC 12A Problems/Problem 17"

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<math>\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004</math>
 
<math>\mathrm{(A)}\ 250\qquad\mathrm{(B)}\ 251\qquad\mathrm{(C)}\ 501\qquad\mathrm{(D)}\ 502\qquad\mathrm{(E)} 1004</math>
  
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==Solution==
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==Solution 1==
 
All positive integers can be expressed as <math>4n</math>, <math>4n+1</math>, <math>4n+2</math>, or <math>4n+3</math>, where <math>n</math> is a nonnegative integer.  
 
All positive integers can be expressed as <math>4n</math>, <math>4n+1</math>, <math>4n+2</math>, or <math>4n+3</math>, where <math>n</math> is a nonnegative integer.  
  
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Since one fourth of the positive integers <math>a_1 \le 2008</math> can be expressed as <math>4n+3</math>, where <math>n</math> is a nonnegative integer, the answer is <math>\frac{1}{4}\cdot 2008 = 502 \Rightarrow D</math>.
 
Since one fourth of the positive integers <math>a_1 \le 2008</math> can be expressed as <math>4n+3</math>, where <math>n</math> is a nonnegative integer, the answer is <math>\frac{1}{4}\cdot 2008 = 502 \Rightarrow D</math>.
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==Solution 2==
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After checking the first few <math>a_n</math> such as <math>1</math>, <math>2</math> through <math>7</math>, we can see that the only <math>a_1</math> that satisfy the conditions are odd numbers that when tripled and added 1 to, are double an odd number. For example, for <math>a_n=3</math>, we notice the sequence yields <math>10</math>, <math>5</math>, and <math>16</math>, a valid sequence.
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So we can set up an equation, <math>3x + 1 = 2(2k - 1)</math> where x is equal to <math>a_1</math>. Rearranging the equation yields <math>(3x + 3)/4 = k</math>. Experimenting yields that every 4th <math>x</math> after 3 creates an integer, and thus satisfies the sequence condition. So the number of valid solutions is equal to <math>2008/4 = 502 \Rightarrow D</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2008|ab=A|num-b=16|num-a=18}}
 
[[Collatz Problem]]
 
[[Collatz Problem]]
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{{MAA Notice}}

Latest revision as of 12:52, 21 December 2020

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 1

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

Solution 2

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 (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

Collatz Problem The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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