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Difference between revisions of "2019 AMC 10B Problems/Problem 25"

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{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #25]] and [[2019 AMC 12B Problems|2019 AMC 12B #23]]}}
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{{duplicate|[[2019 AMC 10B Problems#Problem 25|2019 AMC 10B #25]] and [[2019 AMC 12B Problems#Problem 23|2019 AMC 12B #23]]}}
  
 
==Problem==
 
==Problem==
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<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math>
 
<math>\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75</math>
  
−
==Solution 1 (recursion)==
+
==Solution 1 (Recursion)==
 
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.
 
We can deduce, from the given restrictions, that any valid sequence of length <math>n</math> will start with a <math>0</math> followed by either <math>10</math> or <math>110</math>.
−
Thus we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, where <math>f(n)</math> is the number of valid sequences of length <math>n</math>.
+
Let <math>f(n)</math> be the number of valid (meaning: the sequence contains 0s and 1s, starts and ends with both 0, and there are no two consecutive 0s and no three consecutive 1s) sequences of length <math>n</math>.  
  
−
This is because for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> and the resulting sequence will still satisfy the given conditions.
+
Then we can define a recursive function <math>f(n) = f(n-3) + f(n-2)</math>, with <math>n \ge 3</math> (because otherwise, the sequence would contain only 0s and this is not allowed due to the give conditions).  
  
−
It is easy to find <math>f(5) = 1</math> and <math>f(6) = 2</math> by hand, and then by the recursive formula, we have <math>f(19) = \boxed{\textbf{(C) }65}</math>.
+
We derived the recursive function, since for any valid sequence of length <math>n</math>, you can append either <math>10</math> or <math>110</math> to return to the starting position, 0, and the resulting sequence will still satisfy the given conditions.
 +
 
 +
It is easy to find <math>f(3) = 1</math> since the only possible valid sequence is <math>010</math>. <math>f(4)=1</math> since the only possible valid sequence is <math>0110</math>. <math>f(5)=1</math> since the only possible valid sequence is <math>01010</math>.
 +
 
 +
The recursive sequence is then as follows:
 +
 
 +
<cmath>f(3)=1</cmath>
 +
<cmath>f(4)=1</cmath>
 +
<cmath>f(5) = 1</cmath>
 +
<cmath>f(6) = 1 + 1 = 2</cmath>
 +
<cmath>f(7) = 1 + 1 = 2</cmath>
 +
<cmath>f(8) = 1 + 2 = 3</cmath>
 +
<cmath>f(9) = 2 + 2 = 4</cmath>
 +
<cmath>f(10) = 2 + 3 = 5</cmath>
 +
<cmath>f(11) = 3 + 4 = 7</cmath>
 +
<cmath>f(12) = 4 + 5 = 9</cmath>
 +
<cmath>f(13) = 5 + 7 = 12</cmath>
 +
<cmath>f(14) = 7 + 9 = 16</cmath>
 +
<cmath>f(15) = 9 + 12 = 21</cmath>
 +
<cmath>f(16) = 12 + 16 = 28</cmath>
 +
<cmath>f(17) = 16 + 21 = 37</cmath>
 +
<cmath>f(18) = 21 + 28 = 49</cmath>
 +
<cmath>f(19) = 28 + 37 = 65</cmath>
 +
 
 +
So, our answer is <math>\boxed{\text{\bf{(C)}  } 65}</math>.
 +
 
 +
 
 +
'''Contributors:'''
 +
 
 +
~Original Author
 +
 
 +
~solasky
 +
 
 +
~BakedPotato66
  
 
==Solution 2 (casework)==
 
==Solution 2 (casework)==
Line 22: Line 55:
 
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.
 
'''Case 2''': two <math>3</math>s and six <math>2</math>s - there are <math>{8\choose2} = 28</math> ways to arrange them.
  
−
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose3} = 35</math> ways to arrange them.
+
'''Case 3''': four <math>3</math>s and three <math>2</math>s - there are <math>{7\choose4} = 35</math> ways to arrange them.
  
 
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.
 
'''Case 4''': six <math>3</math>s - there is only <math>1</math> way to arrange them.
Line 28: Line 61:
 
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.
 
Summing the four cases gives <math>1+28+35+1=\boxed{\textbf{(C) }65}</math>.
  
−
==Solution 3(Simpler)==
+
==Solution 3 (casework and blocks)==
−
Firstly, notice how the ending of a suitable string could be a 010, or a 0110. This information basically gives us the answer, as we know that for any string of length N, the possible solutions, would be a string of length n-2, with 10 appended, and a string of length n-3, with 110 appended.  
+
We can simplify the original problem into a problem where there are <math>2^{17}</math> binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of <math>0</math>s, <math>1</math>s, and <math>11</math>s. Now, we use casework:
−
For example:
+
 
−
The strings of length 4 are 0110, the strings of length 5 are 01010, so to make length 7 we can add a 10 to length 5, 0101010, and then add 110 to the string of length 4, 0110110, these are the only possible strings, since the endings have to be 10, or 110, and the strings of length 5 and 4, work and can be modified to work as a 7, since they end with 0.  
+
'''Case 1''': Alternating 1s and 0s. There is simply 1 way to do this: <math>0101010101010101010</math>.  
 +
Now, we note that there cannot be only one block of <math>11</math> in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of <math>11</math>s this cannot be satisfied. This is true for all odd numbers of <math>11</math> blocks.
 +
 
 +
'''Case 2''': There are 2 <math>11</math> blocks. Using the zeroes in the sequence as dividers, we have a sample as <math>0110110101010101010</math>. We know there are 8 places for <math>11</math>s, which will be filled by <math>1</math>s if the <math>11</math>s don't fill them. This is <math>{8\choose2} = 28</math> ways.
 +
 
 +
'''Case 3''': Four <math>11</math> blocks arranged. Using the same logic as Case 2, we have <math>{7\choose4} = 35</math> ways to arrange four <math>11</math> blocks.
 +
 
 +
'''Case 4''': No single <math>1</math> blocks, only <math>11</math> blocks. There is simply one case for this, which is <math>0110110110110110110</math>.
 +
 
 +
Adding these four cases, we have <math>1+28+35+1=\boxed{\textbf{(C) }65}</math> as our final answer.
 +
 
 +
~Equinox8
 +
 
 +
==Solution 4 (similar to #3)==
 +
Any valid sequence must start with a <math>0</math>. We can then think of constructing a sequence as adding groups of terms to this <math>0</math>, each ending in <math>0</math>. (This is always possible because every valid string ends in <math>0</math>.) For example, we can represent the string <math>01011010110110</math> as: <math>0-10-110-10-110-110</math>.
 +
To not have any consecutive 0s, we must have at least one <math>1</math> before the next <math>0</math>. However, we cannot have three or more <math>1</math>s before the next <math>0</math> because we cannot have three consecutive <math>1</math>s. Consequently, we can only have one or two <math>1</math>s.
 +
 
 +
So we can have the groups: <math>10</math> and <math>110</math>.
 +
 
 +
After the initial <math>0</math>, we have <math>18</math> digits left to fill in the string. Let the number of <math>10</math> blocks be <math>x</math>, and <math>110</math> be <math>y</math>. Then <math>x</math> and <math>y</math> must satisfy <math>2x+3y=18</math>. We recognize this as a Diophantine equation. Taking <math>\pmod{2}</math> yields <math>y=0 \pmod{2}</math>. Since <math>x</math> and <math>y</math> must both be nonnegative, we get the solutions <math>(9, 0)</math>, <math>(6, 2)</math>, <math>(3, 4)</math>, and <math>(0, 6)</math>. We now handle each of these cases separately.
 +
 
 +
<math>(9, 0)</math>: Only one arrangement, namely all <math>10</math>s.
 +
 
 +
<math>(6, 2)</math>: We have 6 groups of <math>11</math>, and <math>2</math> groups of <math>110</math>. This has <math>\binom{6+2}{2}=28</math> cases.
 +
 
 +
<math>(3, 4)</math>: This means we have 3 groups of <math>10</math>, and 4 groups of <math>110</math>. This has <math>\binom{3+4}{3}=35</math> cases.
 +
 
 +
<math>(0, 6)</math>: Only one arrangement, namely all <math>110</math>.
 +
 
 +
Adding these, we have <math>1+28+35+1=65 \longrightarrow \boxed{(C)}</math>.
 +
~Math4Life2020
 +
 
 +
~edited by alpha_2 for spelling and and typos
 +
 
 +
==Solution 5 (Constructive Counting)==
 +
Suppose the number of <math>0</math>s is <math>n</math>. We can construct the sequence in two steps:
 +
 
 +
Step 1: put <math>n-1</math> of <math>1</math>s between the <math>0</math>s;
 +
 
 +
Step 2: put the rest <math>19-n-(n-1)=20-2n</math> of <math>1</math>s in the <math>n-1</math> spots where there is a <math>1</math>. There are <math>\binom{n-1}{20-2n}</math> ways of doing this.
 +
 
 +
Now we find the possible values of <math>n</math>:
 +
 
 +
First of all <math>n+(n-1) \leq 19 \Rightarrow n\leq 10</math> (otherwise there will be two consecutive <math>0</math>s);
 +
 
 +
And secondly <math>20-2n \leq n-1\Rightarrow n\geq 7</math> (otherwise there will be three consecutive <math>1</math>s).
 +
 
 +
Therefore the answer is
 +
<cmath>
 +
\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{\textbf{(C) }65}.
 +
</cmath>
 +
 
 +
~ asops
 +
 
 +
== Solution 6 (Recursion)==
 +
 
 +
For a valid sequence of length <math>n</math>, the sequence must be in the form of <math>01xx...xx10</math>. By removing the <math>01</math> at the start of the sequence and the <math>10</math> at the end of the sequence, there are <math>n-4</math> bits left. The <math>n-4</math> bits left can be in the form of:
 +
<math>0yy...yy0</math>, the whole <math>(n-4)</math> bits are valid sequence, which is <math>f(n-4)</math>
 +
<math>0yy...y01</math>, the <math>(n-5)</math> bits before the last <math>1</math> are valid sequence, which is <math>f(n-5)</math>
 +
<math>10y...yy0</math>, the <math>(n-5)</math> bits after the first <math>1</math> are valid sequence, which is <math>f(n-5)</math>
 +
<math>10y...y01</math>, the <math>(n-6)</math> bits between the first and last <math>1</math> are valid sequence, which is <math>f(n-6)</math>
 +
 
 +
So, <math>f(n) = f(n-4) + 2f(n-5) + f(n-6)</math>
 +
 
 +
We will calculate <math>f(19)</math> by [https://en.wikipedia.org/wiki/Dynamic_programming Dynamic Programming].
 +
 
 +
<math>f(3) = 1</math>
 +
 
 +
<math>f(4) = 1</math>
 +
 
 +
<math>f(5) = 1</math>
 +
 
 +
<math>f(6) = 2</math>
 +
 
 +
<math>f(7) = 2</math>
 +
 
 +
<math>f(8) = 3</math>
 +
 
 +
<math>f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4</math>
 +
 
 +
<math>f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5</math>
 +
 
 +
<math>f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7</math>
 +
 
 +
<math>f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9</math>
  
 +
<math>f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12</math>
  
 +
<math>f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16</math>
  
−
Continue this process until 19 and you get  <math> \boxed{\textbf{(C) }65}</math>.
+
<math>f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21</math>
 +
 
 +
<math>f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28</math>
 +
 
 +
<math>f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37</math>
 +
 
 +
<math>f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49</math>
 +
 
 +
<math>f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{\text{\bf{(C)}  } 65}</math>
 +
 
 +
We can further prove <math>f(n) = f(n-4) + 2f(n-5) + f(n-6)</math> is equivalent to <math>f(n) = f(n-2) + f(n-3)</math>
 +
 
 +
Let <math>k(n) = f(n-2) + f(n-3)</math>
 +
 
 +
<math>k(n-2) = f(n-4) + f(n-5)</math>
 +
 
 +
<math>k(n-3) = f(n-5) + f(n-6)</math>
 +
 
 +
<math>k(n-2)+ k(n-3) = f(n-4) + 2f(n-5) + f(n-6) = f(n)</math>
 +
 
 +
<math>f(n) = k(n-2)+ k(n-3)</math>
 +
 
 +
<math>f(n-2) = k(n-4)+ k(n-5)</math>
 +
 
 +
<math>f(n-3) = k(n-5)+ k(n-6)</math>
 +
 
 +
<math>k(n) = f(n-2) + f(n-3) = k(n-4)+ k(n-5) + k(n-5)+ k(n-6) = k(n-4) + 2k(n-5) + k(n-6)</math>
 +
 
 +
So <math>k(n)</math> is the same as <math>f(n)</math>.
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
==Video Solution==
 
==Video Solution==
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{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}
 
{{AMC12 box|year=2019|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}
−
SUB2PEWDS
+
 
 +
[[Category: Intermediate Combinatorics Problems]]

Revision as of 18:35, 6 February 2022

The following problem is from both the 2019 AMC 10B #25 and 2019 AMC 12B #23, so both problems redirect to this page.

Problem

How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?

$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

Solution 1 (Recursion)

We can deduce, from the given restrictions, that any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Let $f(n)$ be the number of valid (meaning: the sequence contains 0s and 1s, starts and ends with both 0, and there are no two consecutive 0s and no three consecutive 1s) sequences of length $n$.

Then we can define a recursive function $f(n) = f(n-3) + f(n-2)$, with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the give conditions).

We derived the recursive function, since for any valid sequence of length $n$, you can append either $10$ or $110$ to return to the starting position, 0, and the resulting sequence will still satisfy the given conditions.

It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$. $f(4)=1$ since the only possible valid sequence is $0110$. $f(5)=1$ since the only possible valid sequence is $01010$.

The recursive sequence is then as follows:

\[f(3)=1\] \[f(4)=1\] \[f(5) = 1\] \[f(6) = 1 + 1 = 2\] \[f(7) = 1 + 1 = 2\] \[f(8) = 1 + 2 = 3\] \[f(9) = 2 + 2 = 4\] \[f(10) = 2 + 3 = 5\] \[f(11) = 3 + 4 = 7\] \[f(12) = 4 + 5 = 9\] \[f(13) = 5 + 7 = 12\] \[f(14) = 7 + 9 = 16\] \[f(15) = 9 + 12 = 21\] \[f(16) = 12 + 16 = 28\] \[f(17) = 16 + 21 = 37\] \[f(18) = 21 + 28 = 49\] \[f(19) = 28 + 37 = 65\]

So, our answer is $\boxed{\text{\bf{(C)} } 65}$.


Contributors:

~Original Author

~solasky

~BakedPotato66

Solution 2 (casework)

After any particular $0$, the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$, i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$s and $3$s to get $18$. There are a number of ways to do this:

Case 1: nine $2$s - there is only $1$ way to arrange them.

Case 2: two $3$s and six $2$s - there are ${8\choose2} = 28$ ways to arrange them.

Case 3: four $3$s and three $2$s - there are ${7\choose4} = 35$ ways to arrange them.

Case 4: six $3$s - there is only $1$ way to arrange them.

Summing the four cases gives $1+28+35+1=\boxed{\textbf{(C) }65}$.

Solution 3 (casework and blocks)

We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$s, $1$s, and $11$s. Now, we use casework:

Case 1: Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$. Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.

Case 2: There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$. We know there are 8 places for $11$s, which will be filled by $1$s if the $11$s don't fill them. This is ${8\choose2} = 28$ ways.

Case 3: Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.

Case 4: No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$.

Adding these four cases, we have $1+28+35+1=\boxed{\textbf{(C) }65}$ as our final answer.

~Equinox8

Solution 4 (similar to #3)

Any valid sequence must start with a $0$. We can then think of constructing a sequence as adding groups of terms to this $0$, each ending in $0$. (This is always possible because every valid string ends in $0$.) For example, we can represent the string $01011010110110$ as: $0-10-110-10-110-110$. To not have any consecutive 0s, we must have at least one $1$ before the next $0$. However, we cannot have three or more $1$s before the next $0$ because we cannot have three consecutive $1$s. Consequently, we can only have one or two $1$s.

So we can have the groups: $10$ and $110$.

After the initial $0$, we have $18$ digits left to fill in the string. Let the number of $10$ blocks be $x$, and $110$ be $y$. Then $x$ and $y$ must satisfy $2x+3y=18$. We recognize this as a Diophantine equation. Taking $\pmod{2}$ yields $y=0 \pmod{2}$. Since $x$ and $y$ must both be nonnegative, we get the solutions $(9, 0)$, $(6, 2)$, $(3, 4)$, and $(0, 6)$. We now handle each of these cases separately.

$(9, 0)$: Only one arrangement, namely all $10$s.

$(6, 2)$: We have 6 groups of $11$, and $2$ groups of $110$. This has $\binom{6+2}{2}=28$ cases.

$(3, 4)$: This means we have 3 groups of $10$, and 4 groups of $110$. This has $\binom{3+4}{3}=35$ cases.

$(0, 6)$: Only one arrangement, namely all $110$.

Adding these, we have $1+28+35+1=65 \longrightarrow \boxed{(C)}$. ~Math4Life2020

~edited by alpha_2 for spelling and and typos

Solution 5 (Constructive Counting)

Suppose the number of $0$s is $n$. We can construct the sequence in two steps:

Step 1: put $n-1$ of $1$s between the $0$s;

Step 2: put the rest $19-n-(n-1)=20-2n$ of $1$s in the $n-1$ spots where there is a $1$. There are $\binom{n-1}{20-2n}$ ways of doing this.

Now we find the possible values of $n$:

First of all $n+(n-1) \leq 19 \Rightarrow n\leq 10$ (otherwise there will be two consecutive $0$s);

And secondly $20-2n \leq n-1\Rightarrow n\geq 7$ (otherwise there will be three consecutive $1$s).

Therefore the answer is \[\sum_{n=7}^{10} \binom{n-1}{20-2n} = \binom{6}{6} + \binom{7}{4} + \binom{8}{2} + \binom{9}{0} = \boxed{\textbf{(C) }65}.\]

~ asops

Solution 6 (Recursion)

For a valid sequence of length $n$, the sequence must be in the form of $01xx...xx10$. By removing the $01$ at the start of the sequence and the $10$ at the end of the sequence, there are $n-4$ bits left. The $n-4$ bits left can be in the form of: 

$0yy...yy0$, the whole $(n-4)$ bits are valid sequence, which is $f(n-4)$
$0yy...y01$, the $(n-5)$ bits before the last $1$ are valid sequence, which is $f(n-5)$
$10y...yy0$, the $(n-5)$ bits after the first $1$ are valid sequence, which is $f(n-5)$
$10y...y01$, the $(n-6)$ bits between the first and last $1$ are valid sequence, which is $f(n-6)$

So, $f(n) = f(n-4) + 2f(n-5) + f(n-6)$

We will calculate $f(19)$ by Dynamic Programming.

$f(3) = 1$

$f(4) = 1$

$f(5) = 1$

$f(6) = 2$

$f(7) = 2$

$f(8) = 3$

$f(9) = f(5) + 2 \cdot f(4) + f(3) = 1 + 2 \cdot 1 + 1 = 4$

$f(10) = f(6) + 2 \cdot f(5) + f(4) = 2 + 2 \cdot 1 + 1 = 5$

$f(11) = f(7) + 2 \cdot f(6) + f(5) = 2 + 2 \cdot 1 + 1 = 7$

$f(12) = f(8) + 2 \cdot f(7) + f(6) = 3 + 2 \cdot 2 + 2 = 9$

$f(13) = f(9) + 2 \cdot f(8) + f(7) = 4 + 2 \cdot 3 + 2 = 12$

$f(14) = f(10) + 2 \cdot f(9) + f(8) = 5 + 2 \cdot 4 + 3 = 16$

$f(15) = f(11) + 2 \cdot f(10) + f(9) = 7 + 2 \cdot 5 + 4 = 21$

$f(16) = f(12) + 2 \cdot f(11) + f(10) = 9 + 2 \cdot 7 + 5 = 28$

$f(17) = f(13) + 2 \cdot f(12) + f(11) = 12 + 2 \cdot 9 + 7 = 37$

$f(18) = f(14) + 2 \cdot f(13) + f(12) = 16 + 2 \cdot 12 + 9 = 49$

$f(19) = f(15) + 2 \cdot f(14) + f(13) = 21 + 2 \cdot 16 + 12 = \boxed{\text{\bf{(C)} } 65}$

We can further prove $f(n) = f(n-4) + 2f(n-5) + f(n-6)$ is equivalent to $f(n) = f(n-2) + f(n-3)$

Let $k(n) = f(n-2) + f(n-3)$

$k(n-2) = f(n-4) + f(n-5)$

$k(n-3) = f(n-5) + f(n-6)$

$k(n-2)+ k(n-3) = f(n-4) + 2f(n-5) + f(n-6) = f(n)$

$f(n) = k(n-2)+ k(n-3)$

$f(n-2) = k(n-4)+ k(n-5)$

$f(n-3) = k(n-5)+ k(n-6)$

$k(n) = f(n-2) + f(n-3) = k(n-4)+ k(n-5) + k(n-5)+ k(n-6) = k(n-4) + 2k(n-5) + k(n-6)$

So $k(n)$ is the same as $f(n)$.

~isabelchen

Video Solution

For those who want a video solution: https://youtu.be/VamT49PjmdI

See Also

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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 AMC 10 Problems and Solutions
2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AMC 12 Problems and Solutions

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