Difference between revisions of "2020 AMC 8 Problems/Problem 22"
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− | ==Problem | + | ==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. | ||
− | + | <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 <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> | ||
+ | |||
+ | ==Solution 1== | ||
+ | We start with final output of <math>1</math> and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage: | ||
+ | <cmath>\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}</cmath> | ||
+ | where, for example, <math>2</math> must come from <math>4</math> (as there is no integer <math>n</math> satisfying <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, last set in this sequence contains all the numbers which will lead to number <math>1</math> to end of the <math>6</math>-step process, and sum is <math>1+8+10+64=\boxed{\textbf{(E) }83}</math>. | ||
+ | |||
+ | ==Solution 2 (variant of Solution 1)== | ||
+ | As in Solution 1, we work backwards from <math>1</math>, this time showing the possible cases in a tree diagram: | ||
+ | |||
+ | <asy> | ||
+ | // Upper branches | ||
+ | draw((-6, 1.5)--(-5, 1)--(-3, 1)--(-2,0)--(0, 0)); | ||
+ | draw((-6, 0.5)--(-5, 1)); | ||
+ | // Lower branches | ||
+ | draw((-6, -1.5)--(-5, -1.5)--(-4, -1)--(-3, -1)--(-2, 0)); | ||
+ | draw((-6, -0.5)--(-5, -0.5)--(-4, -1)); | ||
+ | |||
+ | label("$1$", (0, 0), UnFill(0.1mm)); | ||
+ | label("$2$", (-1, 0), UnFill(0.1mm)); | ||
+ | label("$4$", (-2, 0), UnFill(0.1mm)); | ||
+ | |||
+ | // Upper branches | ||
+ | label("$1$", (-3, 1), UnFill(0.1mm)); | ||
+ | label("$2$", (-4, 1), UnFill(0.1mm)); | ||
+ | label("$4$", (-5, 1), UnFill(0.1mm)); | ||
+ | label("$\textbf{8}$", (-6, 1.5), UnFill(0.1mm)); | ||
+ | label("$\textbf{1}$", (-6, 0.5), UnFill(0.1mm)); | ||
+ | |||
+ | // Lower branches | ||
+ | label("$8$", (-3, -1), UnFill(0.1mm)); | ||
+ | label("$16$",(-4, -1), UnFill(0.1mm)); | ||
+ | label("$5$", (-5, -0.5), UnFill(0.1mm)); | ||
+ | label("$32$", (-5, -1.5), UnFill(0.1mm)); | ||
+ | label("$\textbf{10}$", (-6, -0.5), UnFill(0.1mm)); | ||
+ | label("$\textbf{64}$", (-6, -1.5), UnFill(0.1mm)); | ||
+ | </asy> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 3 (algebraic)== | ||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Remark== | ||
+ | This function is known as the Collatz conjecture stating that every counting number (<math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>...</math>) will eventually output the sequence (<math>4</math>, <math>2</math>, <math>1</math>, <math>4</math>, <math>2</math>, <math>1</math>, <math>...</math>) if put through the function enough times. It is an unsolved conjecture but has been tested by brute force for every starting number up to <math>2^{68}</math>. | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/UnVo6jZ3Wnk?si=l0KAetejsLeFS2We&t=4703 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (🚀 Just 2 min 🚀)== | ||
+ | https://youtu.be/fEux4gUZFbM | ||
+ | |||
+ | ~<i>Education, the Study of Everything</i> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Tf_80QNGmsI | ||
+ | |||
+ | Please like and subscribe! | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/ytTRdD-LVlo?t=755 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=tItOXq9kvu4 ~David | ||
+ | |||
+ | ==Video Solution by Mathiscool== | ||
+ | |||
+ | https://www.youtube.com/watch?v=aLq7LrZZGhc | ||
+ | |||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/ZrDCymOhDdI | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solutions== | ||
+ | https://youtu.be/lhDFmiKNPBg ~ The Learning Royal | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/YnwkBZTv5Fw?t=1347 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==Video Solution by STEMbreezy== | ||
+ | https://youtu.be/wq8EUCe5oQU?t=83 | ||
+ | |||
+ | ~STEMbreezy | ||
+ | ==Note== | ||
+ | |||
+ | This problem is related to a famous unsolved problem, the Collatz Conjecture, also known as the Hailstone Problem, which essentially asks whether or not integer <math>n</math>, repeatedly put in the machine arbitrarily many times, will eventually reach <math>1</math>. | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC8 box|year=2020|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:17, 12 December 2023
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (variant of Solution 1)
- 4 Solution 3 (algebraic)
- 5 Remark
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution (🚀 Just 2 min 🚀)
- 8 Video Solution
- 9 Video Solution by OmegaLearn
- 10 Video Solution
- 11 Video Solution by Mathiscool
- 12 Video Solution by WhyMath
- 13 Video Solutions
- 14 Video Solution by Interstigation
- 15 Video Solution by STEMbreezy
- 16 Note
- 17 See also
Problem
When a positive integer is fed into a machine, the output is a number calculated according to the rule shown below.
For example, starting with an input of
the machine will output
Then if the output is repeatedly inserted into the machine five more times, the final output is
When the same
-step process is applied to a different starting value of
the final output is
What is the sum of all such integers
Solution 1
We start with final output of and work backward, taking cares to consider all possible inputs that could have resulted in any particular output. This produces following set of possibilities each stage:
where, for example,
must come from
(as there is no integer
satisfying
), but
could come from
or
(as
, and
is even while
is odd). By construction, last set in this sequence contains all the numbers which will lead to number
to end of the
-step process, and sum is
.
Solution 2 (variant of Solution 1)
As in Solution 1, we work backwards from , this time showing the possible cases in a tree diagram:
The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are ,
,
, and
. Thus the answer is
.
Solution 3 (algebraic)
We begin by finding the inverse of the function that the machine uses. Call the input and the output
. If
is even,
, and if
is odd,
. We can therefore see that
when
is even and
when
is odd. Therefore, starting with
, if
is even,
, and if
is odd,
, but the latter is not valid since
is not actually odd. This means that the 2nd-to-last number in the sequence has to be
. Now, substituting
into the inverse formulae, if
is even,
(which is indeed even), and if
is odd,
, which is not an integer. This means the 3rd-to-last number in the sequence has to be
. Substituting in
, if
is even,
, but if
is odd,
. Both of these are valid solutions, so the 4th-to-last number can be either
or
. If it is
, then by the argument we have just made, the 5th-to-last number has to be
, the 6th-to-last number has to be
, and the 7th-to-last number, which is the first number, must be either
or
. In this way, we have ultimately found two solutions:
and
.
On the other hand, if the 4th-to-last number is , substituting
into the inverse formulae shows that the 5th-to-last number is either
or
, but the latter is not an integer. Substituting
shows that if
is even,
, but if I is odd,
, and both of these are valid. If the 6th-to-last number is
, then the first number must be
, since
is not an integer; if the 6th-to-last number is
then the first number has to be
, as
is not an integer. This means that, in total, there are
solutions for
, specifically,
,
,
, and
, which sum to
.
Remark
This function is known as the Collatz conjecture stating that every counting number (,
,
,
,
) will eventually output the sequence (
,
,
,
,
,
,
) if put through the function enough times. It is an unsolved conjecture but has been tested by brute force for every starting number up to
.
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=l0KAetejsLeFS2We&t=4703
~Math-X
Video Solution (🚀 Just 2 min 🚀)
~Education, the Study of Everything
Video Solution
Please like and subscribe!
Video Solution by OmegaLearn
https://youtu.be/ytTRdD-LVlo?t=755
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=tItOXq9kvu4 ~David
Video Solution by Mathiscool
https://www.youtube.com/watch?v=aLq7LrZZGhc
Video Solution by WhyMath
~savannahsolver
Video Solutions
https://youtu.be/lhDFmiKNPBg ~ The Learning Royal
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=1347
~Interstigation
Video Solution by STEMbreezy
https://youtu.be/wq8EUCe5oQU?t=83
~STEMbreezy
Note
This problem is related to a famous unsolved problem, the Collatz Conjecture, also known as the Hailstone Problem, which essentially asks whether or not integer , repeatedly put in the machine arbitrarily many times, will eventually reach
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.