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2023 AMC 12A Problems/Problem 20

Revision as of 01:15, 10 November 2023 by Cyantist (talk | contribs) (→‎Solution 2)

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

$S_{2023} = 1 + (x_1 + x_2 + 1) + (x_2 + x_3 + 1) + ... + (x_{2021} + x_{2022} + 1) + 1$

$S_{2023} = x_1 + 2(S_{2022} - x_1 - x_{2022}) + 2023 + x_{2022}$

$S_{2023} = 2S_{2022} + 2021$

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

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

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

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

$S_n = 2^{n-1} + 2^{n-2} + ... + 2^1 - (n-1) + S_1$

$S_n = 2^n - n$

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

The problem requires us to find the last digit of $S_{2023}$. We can use modular arithmetic.

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

$2^{2023} - 2023 \equiv 8 - 3\pmod{10}$

$\equiv 5 \pmod{10}$

~cyantist

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

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