Difference between revisions of "2020 AMC 8 Problems/Problem 22"
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==Solution 4 (Basically the same solution but in detail)== | ==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 | + | 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 |
− | means the 5th-to-last number is either 16 or 7 | ||
==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 20:15, 19 November 2020
Contents
[hide]Problem 22
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 see that work, so
.
~yofro
Solution 2
Start with the final output which is and then work backwards, carefully including all the possible inputs that could have resulted in that output. For example, for the number
, if you go backwards, you only get to
, because
is the only input which can lead to an output of
. However, for a number like
for example, both the inputs
and
lead to an output of
. 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.
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 .
~junaidmansuri
Solution 3
The most straightforward solutions is just working backwards with a diagram as shown below:
Hence, the answer is .
-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 and the output
If
is even,
and if
is odd,
This means the inverse formulas are
when
is even and
when
is odd. From here, we can plug in
into both of these equations to find out what values of
lead to an
value of
If
is even,
and if
is odd,
Note that
is not a valid solution, since
is not odd. This means that the second to last number in the sequence has to be
in order for the last number to be
Next, plug in
into each of these equations. If
is even,
and if
is odd,
Once again,
is not valid, since it has to be a positive integer, but
works. This means the 3rd-to-last number in the sequence has to be
Now comes the first split: if
is even,
but if
is odd,
This means the 4th-to-last number can be either
or
If it is
following the same logic from before, the 5th-to-last number has to be
the 6th-to-last number has to be
and the 7th-to-last number, or the first number, has to be either
or
This gives us
solutions:
or
If the 4th-to-last number is
that means the 5th-to-last number is either
or
But
doesn't work, so it has to be
Now we run into another split: if
is even,
but if I is odd,
If the 6th-to-last number is
the 7th-to-last one,
has to be
since
doesn't work, and if the 6th-to-last number is
then
This means that there are
solutions for
and their sum is
-theepiccarrot7
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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