Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 20"

(→‎Solution 3 (Recursion))
m (Undo revision 232893 by Bakedpotato66 (talk))
(Tag: Undo)
 
(37 intermediate revisions by 7 users not shown)
Line 27: Line 27:
 
==Solution 1==
 
==Solution 1==
  
−
First, let <math>R(n)</math> be the sum of the <math>n</math>th row. Now, with some observations and math instinct, we can guess that <math>R(n) = 2^n - n</math>.
+
First, let <math>R(n)</math> be the sum of the <math>n</math>th row. Now, with some observation and math instinct, we can guess that <math>R(n) = 2^n - n</math>.
  
 
Now we try to prove it by induction,
 
Now we try to prove it by induction,
Line 39: Line 39:
 
By definition from the question, the next row is always<math>:</math>  
 
By definition from the question, the next row is always<math>:</math>  
  
−
Double the sum of last row (Imagine ach number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (leftmost and rightmost are just 1)
+
Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1)
  
 
Which gives us <math>:</math>
 
Which gives us <math>:</math>
Line 63: Line 63:
 
S_{2023} &= 2S_{2022} + 2021
 
S_{2023} &= 2S_{2022} + 2021
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 
From this we can establish:
 
From this we can establish:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 89: Line 90:
 
&\equiv \boxed{\textbf{(C) } 5} \pmod{10} \\
 
&\equiv \boxed{\textbf{(C) } 5} \pmod{10} \\
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
~luckuso
+
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 +
 
 +
==Solution 2.b simpler ==
 +
<cmath>\begin{align*}
 +
S_n &= 2S_{n-1} + n-2 \\
 +
S_n + An + B &= 2(S_{n-1} + A(n-1) + B) \\
 +
A = 1 , B = 0 \\
 +
S_n + n &= 2 (S_{n-1} + n-1) = 2^{n-1} (S_1+1)
 +
\end{align*}</cmath>
 +
continue with the rest from solution 2
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==Solution 3 (Recursion)==
 
==Solution 3 (Recursion)==
Line 95: Line 108:
 
Let the sum of the <math>n^{th}</math> row be <math>S_n</math>.
 
Let the sum of the <math>n^{th}</math> row be <math>S_n</math>.
  
−
For each of the <math>n-2</math> non-1 entries in the <math>n^{th}</math> row, they are equal to the sum of the <math>2</math> numbers diagonally above it in the <math>n-1^{th}</math> row plus <math>1</math>. Therefore all <math>n-3</math> non-1 entries in the <math>n-1^{th}</math> appear twice in sum for the <math>n-2</math> non-1 entries in the <math>n^{th}</math> row, the <math>2</math> <math>1</math>s on each end only appear once in the sum for the <math>n-2</math> non-1 entries in the <math>n^{th}</math> row. Additionally, additional <math>1</math>s are placed at each end of the row. Hence, <math>S_n = 2(S_{n-1} - 1) + n-2 + 1 + 1 = 2 S_{n-1} + n - 2</math>
+
For each of the <math>n-2</math> non-1 entries in the <math>n^{th}</math> row, they are equal to the sum of the <math>2</math> numbers diagonally above it in the <math>n-1^{th}</math> row plus <math>1</math>. Therefore all <math>n-3</math> non-1 entries in the <math>n-1^{th}</math> row appear twice in the sum of the <math>n-2</math> non-1 entries in the <math>n^{th}</math> row, the two <math>1</math>s on each end of the <math>n-1^{th}</math> row only appear once in the sum of the <math>n-2</math> non-1 entries in the <math>n^{th}</math> row. Additionally, additional <math>1</math>s are placed at each end of the <math>n^{th}</math> row. Hence, <math>S_n = 2(S_{n-1} - 1) + n-2 + 1 + 1 = 2 S_{n-1} + n - 2</math>
  
−
<cmath>S_n = 2 S_{n-1} + n - 2</cmath>
+
[[File:2023AMC12AP20 1.png|center|700px]]
−
 
 
−
[[File:2023AMC12AP20.png|center|700px]]
 
  
 
<math>S_4 = 1+5+5+1 = 12</math>, <math>S_5 = 1+7+11+7+1 = 27</math>. By using the recursive formula, <math>S_5 = 2 \cdot S_4 + 5-2 = 27</math>
 
<math>S_4 = 1+5+5+1 = 12</math>, <math>S_5 = 1+7+11+7+1 = 27</math>. By using the recursive formula, <math>S_5 = 2 \cdot S_4 + 5-2 = 27</math>
Line 119: Line 130:
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 +
 +
==Solution 4==
 +
Consider Pascal's triangle as the starting point. In the Pascal's triangle depicted below, the sum of the numbers in the <math>n</math>th row is <math>2^{(n-1)}</math>. For the 2023rd row in the Pascal's triangle, the sum of numbers is <math>2^{2022}</math>.
 +
 +
<asy>
 +
size(4.5cm);
 +
label("$1$", (0,0));
 +
label("$1$", (-0.5,-2/3));
 +
label("$1$", (0.5,-2/3));
 +
label("$1$", (-1,-4/3));
 +
label("$2$", (0,-4/3));
 +
label("$1$", (1,-4/3));
 +
label("$1$", (-1.5,-2));
 +
label("$3$", (-0.5,-2));
 +
label("$3$", (0.5,-2));
 +
label("$1$", (1.5,-2));
 +
label("$1$", (-2,-8/3));
 +
label("$4$", (-1,-8/3));
 +
label("$6$", (0,-8/3));
 +
label("$4$", (1,-8/3));
 +
label("$1$", (2,-8/3));
 +
</asy>
 +
 +
For the triangular array of integers in the problem, 1 is added to each interior entry, which propagates to two numbers diagonally below it in the next row, making the sum of numbers in the next row to increase by 2. And the addition of 1 continues to propagate to the next row, which makes the sum of numbers in the row below the next to increase by 4.
 +
 +
Examine the following triangular array of 1's, which indicates the 1's being added to each position of corresponding numbers in the Pascal's triangle.
 +
 +
<asy>
 +
size(4.5cm);
 +
label("$0$", (0,0));
 +
label("$0$", (-0.5,-2/3));
 +
label("$0$", (0.5,-2/3));
 +
label("$0$", (-1,-4/3));
 +
label("$1$", (0,-4/3));
 +
label("$0$", (1,-4/3));
 +
label("$0$", (-1.5,-2));
 +
label("$1$", (-0.5,-2));
 +
label("$1$", (0.5,-2));
 +
label("$0$", (1.5,-2));
 +
label("$0$", (-2,-8/3));
 +
label("$1$", (-1,-8/3));
 +
label("$1$", (0,-8/3));
 +
label("$1$", (1,-8/3));
 +
label("$0$", (2,-8/3));
 +
</asy>
 +
 +
For the 3rd row, 1 is added to the original number in the same position in the Pascal's triangle. And the addition of 1 in the 3rd row makes the sum of numbers in the 4th row to increase by 2, and makes the sum of numbers in the 5th row to increase by 4, and so forth. Therefore, the addition of a 1 in the 3rd row makes the sum of numbers in the 2023rd row to increase by <math>2^{2023-3}=2^{2020}</math>. And similarly, the addition of each 1 in the 4th row makes the number of numbers in the 2023rd row to increase by <math>2^{2023-4}=2^{2019}</math>. The 1's being added between the 3rd and 2022nd rows impact on the sum of numbers in the 2023rd row to increase by <math>2^{2020} + 2 \cdot 2^{2019} + 3 \cdot 2^{2018} + \dots + 2020 \cdot 2^1 = \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})</math>.
 +
 +
Therefore, the sum of numbers in the 2023rd row is the aggregation of the sum of numbers in the 2023rd row in the Pascal's triangle, the impact of addition of 1's between the 3rd row and the 2022nd row, and the addition of 1's on 2021 interior entries in the 2023rd row, which is <math>2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021</math>.
 +
 +
Because <math>2^{2022} = 2^{2020+2} = 16^{505} \cdot 2^2 = 16^{505} \cdot 4</math> and <math>16^k</math> will always ends with 6 as the unit digit, the unit digit of <math>2^{2022}</math> is 4.
 +
 +
For <math>\sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})</math>, we can solve the sum of geometric sequence to be <math>{2^{2022} - 4 - 2020 \cdot 2}</math>, which has 0 for the unit digit.
 +
 +
Therefore, for the sum of numbers in the 2023rd row, which is <math>2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021</math>, its unit digit is <math>4 + 0 + 1 = \boxed{\textbf{(C) } 5}</math>.
 +
 +
~sqroot
 +
 +
==Solution 5 (mod bash)==
 +
 +
Observe that:
 +
<math>S_n = 2{S_{n-1}} + (n-2) \\
 +
= 2{S_{n-1}} + D_{n} \\
 +
= 2{S_{n-1}} + D_{n-1} + 1
 +
</math>
 +
where <math>D_1=-1</math>, and <math>D_n=D_{n-1}+1</math> for <math>n>1</math>.
 +
 +
Make a table of values, <math>(S_n, D_n) \mod 2</math> and <math>(S_n, D_n) \mod 5</math>, a until the cycle completes for each. (<math>\mod 2</math> has period <math>1</math>; <math>\mod 5</math> has period <math>20</math>).
 +
( In both cases, there is no "head" that is not part of the cycle.)
 +
 +
Mod 2:
 +
<math>S_{2023} \equiv S_{2023 \mod 2}  \equiv  S_1  = (1\cdot 2 +0)\cdot 0 + 1 \equiv 1 \pmod 2</math>
 +
 +
Mod 5:
 +
<math>S_{2023} \equiv S_{2023 \mod 20}  \equiv  S_3 \equiv (1\cdot 2 +0)\cdot 2 + 1 \equiv 5 \pmod 5</math>
 +
 +
Thus <math>S_{2023} \mod 10 = \boxed{\textbf{(C) } 5}</math>.
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==

Latest revision as of 20:44, 9 November 2024

Problem

Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.

[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]

Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?

$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution 1

First, let $R(n)$ be the sum of the $n$th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$.

Now we try to prove it by induction,

$R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case)

$R(k) = 2^k - k$

$R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$

By definition from the question, the next row is always$:$

Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1)

Which gives us $:$

$2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$

Hence, proven

Last, simply substitute $n = 2023$, we get $R(2023) = 2^{2023} - 2023$

Last digit of $2^{2023}$ is $8$, $8-3 = \boxed{\textbf{(C) } 5}$

~lptoggled

Solution 2

Let the sum of the numbers in row $2022$ be $S_{2022}$. Let each number in row $2022$ be $x_i$ where $1 \leq i \leq 2022$.

Then \begin{align*} S_{2023}&=1+(x_1+x_2+1)+(x_2+x_3+1)+...+(x_{2021}+x_{2022}+ 1)+1 \\ S_{2023}&=x_1+2(S_{2022}-x_1-x_{2022})+2023+x_{2022} \\ S_{2023} &= 2S_{2022} + 2021 \end{align*}

From this we can establish: \begin{align*} S_n &= 2S_{n-1} + n-2 \\ S_{n-1} &= 2S_{n-2} + n-3 \\ S_n - S_{n-1} &= 2S_{n-1} - 2S_{n-2} + 1 \end{align*}

Let $B_{n} = S_n - S_{n-1}$

From this we have: \begin{align*} B_n + 1 &= 2(B_{n-1} + 1) \\ B_n &= 2^{n-1} - 1 \\ S_n - S_{n-1} &= 2^{n-1} - 1 \\ S_n &= 2^{n-1} + 2^{n-2} + ... + 2^1 - (n-1) + S_1 \\ S_n &= 2^n - n \\ S_{2023} &= 2^{2023} - 2023 \end{align*}

The problem requires us to find the last digit of $S_{2023}$. We can use modular arithmetic. \begin{align*} S_{2023} &= 2^{2023} - 2023 \\ 2^{2023} - 2023 &\equiv 8 - 3\pmod{10} \\ &\equiv \boxed{\textbf{(C) } 5} \pmod{10} \\ \end{align*}

~luckuso

Solution 2.b simpler

\begin{align*} S_n &= 2S_{n-1} + n-2 \\ S_n + An + B &= 2(S_{n-1} + A(n-1) + B) \\ A = 1 , B = 0 \\ S_n + n &= 2 (S_{n-1} + n-1) = 2^{n-1} (S_1+1) \end{align*} continue with the rest from solution 2

~luckuso

Solution 3 (Recursion)

Let the sum of the $n^{th}$ row be $S_n$.

For each of the $n-2$ non-1 entries in the $n^{th}$ row, they are equal to the sum of the $2$ numbers diagonally above it in the $n-1^{th}$ row plus $1$. Therefore all $n-3$ non-1 entries in the $n-1^{th}$ row appear twice in the sum of the $n-2$ non-1 entries in the $n^{th}$ row, the two $1$s on each end of the $n-1^{th}$ row only appear once in the sum of the $n-2$ non-1 entries in the $n^{th}$ row. Additionally, additional $1$s are placed at each end of the $n^{th}$ row. Hence, $S_n = 2(S_{n-1} - 1) + n-2 + 1 + 1 = 2 S_{n-1} + n - 2$

2023AMC12AP20 1.png

$S_4 = 1+5+5+1 = 12$, $S_5 = 1+7+11+7+1 = 27$. By using the recursive formula, $S_5 = 2 \cdot S_4 + 5-2 = 27$

\[S_n = 2 S_{n-1} + n - 2\] \[S_n + n = 2 S_{n-1} + 2n - 2\] \[S_n + n = 2( S_{n-1} + n - 1)\] $S_n + n$ is a geometric sequence by a ratio of $2$

$\because \quad S_1 = 1$

$\therefore \quad S_n + n = 2^n$, $S_n = 2^n - n$

$S_{2023} = 2^{2023} - 2023$

The unit digit of powers of $2$ is periodic by a cycle of $4$ digits: $2$, $4$, $8$, $6$. $2023 = 3 \mod 4$, the unit digit of $2^{2023}$ is $8$.

Therefore, the unit digit of $S_{2023}$ is $8-3 = \boxed{\textbf{(C) } 5}$

~isabelchen

Solution 4

Consider Pascal's triangle as the starting point. In the Pascal's triangle depicted below, the sum of the numbers in the $n$th row is $2^{(n-1)}$. For the 2023rd row in the Pascal's triangle, the sum of numbers is $2^{2022}$.

[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$2$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$3$", (-0.5,-2)); label("$3$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$4$", (-1,-8/3)); label("$6$", (0,-8/3)); label("$4$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]

For the triangular array of integers in the problem, 1 is added to each interior entry, which propagates to two numbers diagonally below it in the next row, making the sum of numbers in the next row to increase by 2. And the addition of 1 continues to propagate to the next row, which makes the sum of numbers in the row below the next to increase by 4.

Examine the following triangular array of 1's, which indicates the 1's being added to each position of corresponding numbers in the Pascal's triangle.

[asy] size(4.5cm); label("$0$", (0,0)); label("$0$", (-0.5,-2/3)); label("$0$", (0.5,-2/3)); label("$0$", (-1,-4/3)); label("$1$", (0,-4/3)); label("$0$", (1,-4/3)); label("$0$", (-1.5,-2)); label("$1$", (-0.5,-2)); label("$1$", (0.5,-2)); label("$0$", (1.5,-2)); label("$0$", (-2,-8/3)); label("$1$", (-1,-8/3)); label("$1$", (0,-8/3)); label("$1$", (1,-8/3)); label("$0$", (2,-8/3)); [/asy]

For the 3rd row, 1 is added to the original number in the same position in the Pascal's triangle. And the addition of 1 in the 3rd row makes the sum of numbers in the 4th row to increase by 2, and makes the sum of numbers in the 5th row to increase by 4, and so forth. Therefore, the addition of a 1 in the 3rd row makes the sum of numbers in the 2023rd row to increase by $2^{2023-3}=2^{2020}$. And similarly, the addition of each 1 in the 4th row makes the number of numbers in the 2023rd row to increase by $2^{2023-4}=2^{2019}$. The 1's being added between the 3rd and 2022nd rows impact on the sum of numbers in the 2023rd row to increase by $2^{2020} + 2 \cdot 2^{2019} + 3 \cdot 2^{2018} + \dots + 2020 \cdot 2^1 = \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})$.

Therefore, the sum of numbers in the 2023rd row is the aggregation of the sum of numbers in the 2023rd row in the Pascal's triangle, the impact of addition of 1's between the 3rd row and the 2022nd row, and the addition of 1's on 2021 interior entries in the 2023rd row, which is $2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021$.

Because $2^{2022} = 2^{2020+2} = 16^{505} \cdot 2^2 = 16^{505} \cdot 4$ and $16^k$ will always ends with 6 as the unit digit, the unit digit of $2^{2022}$ is 4.

For $\sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})$, we can solve the sum of geometric sequence to be ${2^{2022} - 4 - 2020 \cdot 2}$, which has 0 for the unit digit.

Therefore, for the sum of numbers in the 2023rd row, which is $2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021$, its unit digit is $4 + 0 + 1 = \boxed{\textbf{(C) } 5}$.

~sqroot

Solution 5 (mod bash)

Observe that: $S_n = 2{S_{n-1}} + (n-2) \\ = 2{S_{n-1}} + D_{n} \\ = 2{S_{n-1}} + D_{n-1} + 1$ where $D_1=-1$, and $D_n=D_{n-1}+1$ for $n>1$.

Make a table of values, $(S_n, D_n) \mod 2$ and $(S_n, D_n) \mod 5$, a until the cycle completes for each. ($\mod 2$ has period $1$; $\mod 5$ has period $20$). ( In both cases, there is no "head" that is not part of the cycle.)

Mod 2: $S_{2023} \equiv S_{2023 \mod 2} \equiv S_1 = (1\cdot 2 +0)\cdot 0 + 1 \equiv 1 \pmod 2$

Mod 5: $S_{2023} \equiv S_{2023 \mod 20} \equiv S_3 \equiv (1\cdot 2 +0)\cdot 2 + 1 \equiv 5 \pmod 5$

Thus $S_{2023} \mod 10 = \boxed{\textbf{(C) } 5}$.

Video Solution 1 by OmegaLearn

https://youtu.be/MnR5VXAXDeU

Video Solution

https://youtu.be/EVHnJFOW4Ys

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_20&oldid=232894"