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"

(→‎Problem)
m (→‎Solution)
Line 52: Line 52:
 
<math>B_n = 2^{n-1} - 1</math>
 
<math>B_n = 2^{n-1} - 1</math>
  
−
==Solution==
+
==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 observations and math instinct, we can guess that <math>R(n) = 2^n - n</math>.

Revision as of 00:56, 10 November 2023

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 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*} $S_{2023} &= 1 + (x_1 + x_2 + 1) + (x_2 + x_3 + 1) + ... + (x_{2021} + x_{2022} + 1) + 1$ (Error compiling LaTeX. Unknown error_msg)\\

$S_{2023} &= x_1 + 2(S_{2022} - x_1 - x_{2022}) + 2023 + x_{2022}$ (Error compiling LaTeX. Unknown error_msg)\\

$S_{2023} &= 2S_{2022} + 2021$ (Error compiling LaTeX. Unknown error_msg) \end{align*} From this we can establish 2 equations:

$S_n = 2S_{n-1} + n-2$

$S_{n-1} = 2S_{n-2} + n-3$

$S_n - S_{n-1} = 2S_{n-1} - 2S_{n-2} + 1$

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

$B_n + 1 = 2(B_{n-1} + 1)$

$B_n = 2^{n-1} - 1$

Solution 1

First, let $R(n)$ be the sum of the $n$th row. Now, with some observations 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 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)

Which gives us $:$

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

Hence, proven (you can work it backwards without induction & guessing)

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

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=201737"