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

Revision as of 09:03, 19 November 2020

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:

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

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=I/2, and if I is odd, O=3I+1. This means the inverse formulas are I=2O when I is even and I=(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=1/3. Once again, 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 7/3. 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 31/2 doesn't work, and if the 6th-to-last number is 5, N=10. This means that there are 4 solutions for N, 1, 8, 10, and 64, and their sum is 83. -theepiccarrot7

Revision as of 09:03, 19 November 2020 (view source) Volcanogobbler (talk \| contribs) (→‎Solution 3) ← Older edit		Revision as of 09:03, 19 November 2020 (view source) Volcanogobbler (talk \| contribs) (→‎Solution 3) Newer edit →
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	[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC8yLzc0MjY0Yjc2NDcyMDgwNTBlZDlhMDlmNjNmMjI1Y2RhNmUwODkyLnBuZw==&rn=MjAyMCBBTUMgOCBTb2x1dGlvbiAyMi5wbmc=[/img]		[img]https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZC8yLzc0MjY0Yjc2NDcyMDgwNTBlZDlhMDlmNjNmMjI1Y2RhNmUwODkyLnBuZw==&rn=MjAyMCBBTUMgOCBTb2x1dGlvbiAyMi5wbmc=[/img]
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	Hence, the answer is <math>1+8+64+10=\textbf{(E) }83</math>.		Hence, the answer is <math>1+8+64+10=\textbf{(E) }83</math>.

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