2023 AMC 12A Problems/Problem 20

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*}

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

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 $8-3 = \boxed{\textbf{(C) } 5}$

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

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