2020 AMC 8 Problems/Problem 22

Problem 22

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 see that $1,8,10,64$ work, so $\boxed{\textbf{(E) }83}$ . ~yofro

Solution 2

Start with the final output which is $1$ and then work backwards, carefully including all the possible inputs that could have resulted in that output. For example, for the number $2$ , if you go backwards, you only get to $4$ , because $4$ is the only input which can lead to an output of $2$ . However, for a number like $16$ for example, both the inputs $5$ and $32$ lead to an output of $16$ . 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.

$\{1\}\leftarrow\{2\}\leftarrow\{4\}\leftarrow\{1,8\}\leftarrow\{2,16\}\leftarrow\{4,5,32\}\leftarrow\{1,8,10,64\}$

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 $1+8+10+64=83\implies\boxed{\textbf{(E) }83}$ .
~junaidmansuri

Solution 3

The most straightforward solutions is just working backwards with a diagram as shown below:

https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC8yLzc0MjY0Yjc2NDcyMDgwNTBlZDlhMDlmNjNmMjI1Y2RhNmUwODkyLnBuZw==&rn=MjAyMCBBTUMgOCBTb2x1dGlvbiAyMi5wbmc=

Hence, the answer is $1+8+64+10=\textbf{(E) }83$ .

-franzliszt

Solution 4 (Basically the same solution but in detail)

To solve this, we can work backward. First, we find 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, $O=3I+1.$ This means the inverse formulas are $I=2O$ when $I$ is even and $I=\frac{O-1}{3}$ when $I$ is odd. From here, we can plug in $1$ into both of these equations to find out what values of $I$ lead to an $O$ value of $1.$ If $I$ is even, $I=2,$ and if $I$ is odd, $I=0.$ Note that $I=0$ is not a valid solution, since $0$ is not odd. This means that the second to last number in the sequence has to be $2$ in order for the last number to be $1.$ Next, plug in $2$ into each of these equations. If $I$ is even, $I=4,$ and if $I$ is odd, $I=\frac{1}{3}.$ Once again, $\frac{1}{3}$ is not valid, since it has to be a positive integer, but $4$ works. This means the 3rd-to-last number in the sequence has to be $4.$ Now comes the first split: if $I$ is even, $I=8,$ but if $I$ is odd, $I=1.$ This means the 4th-to-last number can be either $1$ or $8.$ If it is $1,$ following the same logic from before, the 5th-to-last number has to be $2,$ the 6th-to-last number has to be $4,$ and the 7th-to-last number, or the first number, has to be either $1$ or $8.$ This gives us $2$ solutions: $N=1,$ or $N=8.$ If the 4th-to-last number is $8,$ that means the 5th-to-last number is either $16$ or $\frac{7}{3}.$ But $\frac{7}{3}$ doesn't work, so it has to be $16.$ Now we run into another split: if $I$ is even, $I=32,$ but if I is odd, $I=5.$ If the 6th-to-last number is $32,$ the 7th-to-last one, $N,$ has to be $64,$ since $\frac{31}{2}$ doesn't work, and if the 6th-to-last number is $5,$ then $N=10.$ This means that there are $4$ solutions for $N:$ $1, 8, 10, and 64,$ and their sum is $83.$ -theepiccarrot7

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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