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Difference between revisions of "2020 AMC 8 Problems/Problem 22"

(→‎Solution 4 (Basically the same solution but in detail))
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==Problem 22==
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==Problem==
 
When a positive integer <math>N</math> is fed into a machine, the output is a number calculated according to the rule shown below.
 
When a positive integer <math>N</math> is fed into a machine, the output is a number calculated according to the rule shown below.
  
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For example, starting with an input of <math>N=7,</math> the machine will output <math>3 \cdot 7 +1 = 22.</math> Then if the output is repeatedly inserted into the machine five more times, the final output is <math>26.</math><cmath>7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26</cmath>When the same <math>6</math>-step process is applied to a different starting value of <math>N,</math> the final output is <math>1.</math> What is the sum of all such integers <math>N?</math><cmath>N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1</cmath>
 
For example, starting with an input of <math>N=7,</math> the machine will output <math>3 \cdot 7 +1 = 22.</math> Then if the output is repeatedly inserted into the machine five more times, the final output is <math>26.</math><cmath>7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26</cmath>When the same <math>6</math>-step process is applied to a different starting value of <math>N,</math> the final output is <math>1.</math> What is the sum of all such integers <math>N?</math><cmath>N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1</cmath>
 
<math>\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83</math>
 
<math>\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83</math>
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==Solution 1==
 
==Solution 1==
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We see that <math>1,8,10,64</math> work, so <math>\boxed{\textbf{(E) }83}</math>.
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We start with the final output of <math>1</math> and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage:
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~yofro
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<cmath>\{1\}\leftarrow\{2\}\leftarrow\{4\}\leftarrow\{1,8\}\leftarrow\{2,16\}\leftarrow\{4,5,32\}\leftarrow\{1,8,10,64\}</cmath>
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where, for example, <math>2</math> must come from <math>4</math> (as there is no integer <math>n</math> with <math>3n+1=2</math>), but <math>16</math> could come from <math>32</math> or <math>5</math> (as <math>\frac{32}{2} = 3 \cdot 5 + 1 = 16</math>, and <math>32</math> is even while <math>5</math> is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number <math>1</math> at the end of the <math>6</math>-step process, and their sum is <math>1+8+10+64=\boxed{\textbf{(E) }83}</math>.
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==Solution 2==
 
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Start with the final output which is <math>1</math> and then work backwards, carefully including all the possible inputs that could have resulted in that output. For example, for the number <math>2</math>, if you go backwards, you only get to <math>4</math>, because <math>4</math> is the only input which can lead to an output of <math>2</math>. However, for a number like <math>16</math> for example, both the inputs <math>5</math> and <math>32</math> lead to an output of <math>16</math>. A nice way to draw this out is to make a tree diagram but one can also make a series of sets which contain all the possible inputs up to that point.<br><br>
 
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<math>\{1\}\leftarrow\{2\}\leftarrow\{4\}\leftarrow\{1,8\}\leftarrow\{2,16\}\leftarrow\{4,5,32\}\leftarrow\{1,8,10,64\}</math><br><br>
 
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The last set in this sequence contains all the numbers which will lead to the number 1 after the 6-step process is repeated. The sum of these numbers is <math>1+8+10+64=83\implies\boxed{\textbf{(E) }83}</math>.<br>
 
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~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]
 
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==Solution 3==
 
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The most straightforward solutions is just working backwards with a diagram as shown below:
 
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https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC8yLzc0MjY0Yjc2NDcyMDgwNTBlZDlhMDlmNjNmMjI1Y2RhNmUwODkyLnBuZw==&rn=MjAyMCBBTUMgOCBTb2x1dGlvbiAyMi5wbmc=
 
  
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Hence, the answer is <math>1+8+64+10=\textbf{(E) }83</math>.
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==Solution 2 (variant of Solution 1)==
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As in Solution 1, we work backwards from <math>1</math>, this time showing the possible cases in a tree diagram:
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[[File:Prob22-diagram.png|middle|center]]
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The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are <math>1</math>, <math>8</math>, <math>64</math>, and <math>10</math>. Thus the answer is <math>1+8+64+10=\boxed{\textbf{(E) }83}</math>.
  
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-franzliszt
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==Solution 3 (algebraic)==
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We begin by finding the inverse of the function that the machine uses. Call the input <math>I</math> and the output <math>O</math>. If <math>I</math> is even, <math>O=\frac{I}{2}</math>, and if <math>I</math> is odd, <math>O=3I+1</math>. We can therefore see that <math>I=2O</math> when <math>I</math> is even and <math>I=\frac{O-1}{3}</math> when <math>I</math> is odd. Therefore, starting with <math>1</math>, if <math>I</math> is even, <math>I=2</math>, and if <math>I</math> is odd, <math>I=0</math>, but the latter is not valid since <math>0</math> is not actually odd. This means that the 2nd-to-last number in the sequence has to be <math>2</math>. Now, substituting <math>2</math> into the inverse formulae, if <math>I</math> is even, <math>I=4</math> (which is indeed even), and if <math>I</math> is odd, <math>I=\frac{1}{3}</math>, which is not an integer. This means the 3rd-to-last number in the sequence has to be <math>4</math>. Substituting in <math>4</math>, if <math>I</math> is even, <math>I=8</math>, but if <math>I</math> is odd, <math>I=1</math>. Both of these are valid solutions, so the 4th-to-last number can be either <math>1</math> or <math>8</math>. If it is <math>1</math>, then by the argument we have just made, the 5th-to-last number has to be <math>2</math>, the 6th-to-last number has to be <math>4</math>, and the 7th-to-last number, which is the first number, must be either <math>1</math> or <math>8</math>. In this way, we have ultimately found two solutions: <math>N=1</math> and <math>N=8</math>.
  
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==Solution 4 (Basically the same solution but in detail)==
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On the other hand, if the 4th-to-last number is <math>8</math>, substituting <math>8</math> into the inverse formulae shows that the 5th-to-last number is either <math>16</math> or <math>\frac{7}{3}</math>, but the latter is not an integer. Substituting <math>16</math> shows that if <math>I</math> is even, <math>I=32</math>, but if I is odd, <math>I=5</math>, and both of these are valid. If the 6th-to-last number is <math>32</math>, then the first number must be <math>64</math>, since <math>\frac{31}{3}</math> is not an integer; if the 6th-to-last number is <math>5,</math> then the first number has to be <math>10</math>, as <math>\frac{4}{3}</math> is not an integer. This means that, in total, there are <math>4</math> solutions for <math>N</math>, specifically, <math>1</math>, <math>8</math>, <math>10</math>, and <math>64</math>, which sum to <math>\boxed{\textbf{(E) }83}</math>.
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To solve this, we can work backward. First, we find the inverse of the function that the machine uses. Call the input <math>I</math> and the output <math>O.</math> If <math>I</math> is even, <math>O=\frac{I}{2},</math> and if <math>I</math> is odd, <math>O=3I+1.</math> This means the inverse formulas are <math>I=2O</math> when <math>I</math> is even and <math>I=\frac{O-1}{3}</math> when <math>I</math> is odd. From here, we can plug in <math>1</math> into both of these equations to find out what values of <math>I</math> lead to an <math>O</math> value of <math>1.</math> If <math>I</math> is even, <math>I=2,</math> and if <math>I</math> is odd, <math>I=0.</math> Note that <math>I=0</math> is not a valid solution, since <math>0</math> is not odd. This means that the second to last number in the sequence has to be <math>2</math> in order for the last number to be <math>1.</math> Next, plug in <math>2</math> into each of these equations. If <math>I</math> is even, <math>I=4,</math> and if <math>I</math> is odd, <math>I=\frac{1}{3}.</math> Once again, <math>\frac{1}{3}</math> is not valid, since it has to be a positive integer, but <math>4</math> works. This means the 3rd-to-last number in the sequence has to be <math>4.</math> Now comes the first split: if <math>I</math> is even, <math>I=8,</math> but if <math>I</math> is odd, <math>I=1.</math> This means the 4th-to-last number can be either <math>1</math> or <math>8.</math> If it is <math>1,</math> following the same logic from before, the 5th-to-last number has to be <math>2,</math> the 6th-to-last number has to be <math>4,</math> and the 7th-to-last number, or the first number, has to be either <math>1</math> or <math>8.</math> This gives us <math>2</math> solutions: <math>N=1,</math> or <math>N=8.</math> If the 4th-to-last number is <math>8,</math> that means the 5th-to-last number is either <math>16</math> or <math>\frac{7}{3}.</math> But <math>\frac{7}{3}</math> doesn't work, so it has to be <math>16.</math> Now we run into another split: if <math>I</math> is even, <math>I=32,</math> but if I is odd, <math>I=5.</math> If the 6th-to-last number is <math>32,</math> the 7th-to-last one, <math>N,</math> has to be <math>64,</math> since <math>\frac{31}{2}</math> doesn't work, and if the 6th-to-last number is <math>5,</math> then <math>N=10.</math> This means that there are <math>4</math> solutions for <math>N:</math> <math>1, 8, 10, and 64,</math> and their sum is <math>83.</math> -theepiccarrot7
 
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=21|num-a=23}}
 
{{AMC8 box|year=2020|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:49, 20 November 2020

Problem

When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.

[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] For example, starting with an input of $N=7,$ the machine will output $3 \cdot 7 +1 = 22.$ Then if the output is repeatedly inserted into the machine five more times, the final output is $26.$\[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\]When the same $6$-step process is applied to a different starting value of $N,$ the final output is $1.$ What is the sum of all such integers $N?$\[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\] $\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

Solution 1

We start with the final output of $1$ and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage: \[\{1\}\leftarrow\{2\}\leftarrow\{4\}\leftarrow\{1,8\}\leftarrow\{2,16\}\leftarrow\{4,5,32\}\leftarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ with $3n+1=2$), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$, and $32$ is even while $5$ is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number $1$ at the end of the $6$-step process, and their sum is $1+8+10+64=\boxed{\textbf{(E) }83}$.

Solution 2 (variant of Solution 1)

As in Solution 1, we work backwards from $1$, this time showing the possible cases in a tree diagram:

Prob22-diagram.png

The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are $1$, $8$, $64$, and $10$. Thus the answer is $1+8+64+10=\boxed{\textbf{(E) }83}$.

Solution 3 (algebraic)

We begin by finding the inverse of the function that the machine uses. Call the input $I$ and the output $O$. If $I$ is even, $O=\frac{I}{2}$, and if $I$ is odd, $O=3I+1$. We can therefore see that $I=2O$ when $I$ is even and $I=\frac{O-1}{3}$ when $I$ is odd. Therefore, starting with $1$, if $I$ is even, $I=2$, and if $I$ is odd, $I=0$, but the latter is not valid since $0$ is not actually odd. This means that the 2nd-to-last number in the sequence has to be $2$. Now, substituting $2$ into the inverse formulae, if $I$ is even, $I=4$ (which is indeed even), and if $I$ is odd, $I=\frac{1}{3}$, which is not an integer. This means the 3rd-to-last number in the sequence has to be $4$. Substituting in $4$, if $I$ is even, $I=8$, but if $I$ is odd, $I=1$. Both of these are valid solutions, so the 4th-to-last number can be either $1$ or $8$. If it is $1$, then by the argument we have just made, the 5th-to-last number has to be $2$, the 6th-to-last number has to be $4$, and the 7th-to-last number, which is the first number, must be either $1$ or $8$. In this way, we have ultimately found two solutions: $N=1$ and $N=8$.

On the other hand, if the 4th-to-last number is $8$, substituting $8$ into the inverse formulae shows that the 5th-to-last number is either $16$ or $\frac{7}{3}$, but the latter is not an integer. Substituting $16$ shows that if $I$ is even, $I=32$, but if I is odd, $I=5$, and both of these are valid. If the 6th-to-last number is $32$, then the first number must be $64$, since $\frac{31}{3}$ is not an integer; if the 6th-to-last number is $5,$ then the first number has to be $10$, as $\frac{4}{3}$ is not an integer. This means that, in total, there are $4$ solutions for $N$, specifically, $1$, $8$, $10$, and $64$, which sum to $\boxed{\textbf{(E) }83}$.

See also

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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 AJHSME/AMC 8 Problems and Solutions

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