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

Revision as of 15:36, 13 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 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 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, $S_n = 2(S_{n-1} - 1) + n-2 + 1 + 1 = 2 S_{n-1} + n - 2$

$S_4 = 1+5+5+1 = 12$ , $S_5 = 1+7+11+7+1 = 27$ . 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 Nth 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 diagonally-below numbers in the next row, and the impact of the 1 continues to the next row and so forth. Those 1's being added to a number in the Pascal's triangle between the 3rd row and the 2022nd row eventually propagate to the numbers in the 2023rd row twice. Therefore, the sum of numbers in the 2023rd row of this triangular array of integers will be the sum of numbers in the 2023rd row in the Pascal's triangle, plus two times all the 1's that have been added between the 3rd row and the 2022nd row, plus all the 1's of the 2021 interior entries in the 2023rd row.

Examine the following triangular array of 1's, there are 2020 rows between the 3rd and 2022nd rows, therefore the total number of 1's between the 3rd and 2022nd rows is ${2020 \cdot 2021/2}$ . $[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]$

The sum of numbers in the 2023rd row will be $2^{2022} + {2020 \cdot 2021/2} + 2021$ . For $2^{2022}$ , the unit digit is the same as that of $2^2$ , which is 4. Therefore, the unit digit for the sum of numbers in the 2023rd row is $\boxed{\textbf{(C) } 5}$ .

~sqroot

Video Solution 1 by OmegaLearn

https://youtu.be/MnR5VXAXDeU

Video Solution

https://youtu.be/EVHnJFOW4Ys

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

@@ Line 140: / Line 140: @@
 </asy>
-For the triangular array of integers in the problem, 1 is added to each interior entry, which propagates to two numbers in the next row. Those 1's being added to a number in the Pascal's triangle between the 3rd row and the 2022nd row eventually propagate to the numbers in the 2023rd row twice. Therefore, the sum of numbers in the 2023rd row of this triangular array of integers will be the sum of numbers in the 2023rd row in the Pascal's triangle, plus two times all the 1's that have been added between the 3rd row and the 2022nd row, plus all the 1's of the 2021 interior entries in the 2023rd row.
+For the triangular array of integers in the problem, 1 is added to each interior entry, which propagates to two diagonally-below numbers in the next row, and the impact of the 1 continues to the next row and so forth. Those 1's being added to a number in the Pascal's triangle between the 3rd row and the 2022nd row eventually propagate to the numbers in the 2023rd row twice. Therefore, the sum of numbers in the 2023rd row of this triangular array of integers will be the sum of numbers in the 2023rd row in the Pascal's triangle, plus two times all the 1's that have been added between the 3rd row and the 2022nd row, plus all the 1's of the 2021 interior entries in the 2023rd row.
 Examine the following triangular array of 1's, there are 2020 rows between the 3rd and 2022nd rows, therefore the total number of 1's between the 3rd and 2022nd rows is <math>{2020 \cdot 2021/2}</math>.

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

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