AMC 12A 2023 (Problem 20)

Rows

1,2,3,4,

and

5

of a triangular array of integers are shown below:

\begin{array}{ccccccccc}&&&&1&&&&\\[6pt]&&&1&&1&&&\\[6pt]&&1&&3&&1&&\\[6pt]&1&&5&&5&&1&\\[6pt]1&&7&&11&&7&&1\end{array}

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 digit of the sum of the

2023

numbers in the

2023

rd row?

\text{(A)}\;1

\text{(B)}\;3

\text{(C)}\;5

\text{(D)}\;7

\text{(E)}\;9

\frac{0}{2^2}+\frac{1}{2^3}+\frac{2}{2^4}+\frac{3}{2^5}+\cdots+\frac{n-3}{2^{n-1}}+\frac{n-2}{2^n}

\begin{array}{cccccccccccl}\frac1{2^3}&+&\frac1{2^4}&+&\frac1{2^5}&+&\cdots&+&\frac1{2^{n-1}}&+&\frac1{2^n}&\quad\Rightarrow\quad \frac1{2^3}\left(1+\frac12+\frac1{2^2}+\cdots+\frac1{2^{n-3}}\right)=\frac1{2^2}-\frac1{2^n}\\[10pt]&&\frac1{2^4}&+&\frac1{2^5}&+&\cdots&+&\frac1{2^{n-1}}&+&\frac1{2^n}&\quad\Rightarrow\quad \frac1{2^4}\left(1+\frac12+\cdots+\frac1{2^{n-4}}\right)=\frac1{2^3}-\frac1{2^n}\\[10pt]&&&&\frac1{2^5}&+&\cdots&+&\frac1{2^{n-1}}&+&\frac1{2^n}&\quad\Rightarrow\quad \frac1{2^4}-\frac1{2^n}\\[8pt]&&&&&&\vdots&&&&&\\[8pt]&&&&&&\frac1{2^{n-1}}&+&\frac1{2^n}&\quad\Rightarrow\quad \frac1{2^{n-2}}-\frac1{2^n}\\[10pt]&&&&&&&&\frac1{2^n}&\quad\Rightarrow\quad \frac1{2^{n-1}}-\frac1{2^n}\end{array}

\Rightarrow\left(\frac1{2^2}-\frac1{2^n}\right)+\left(\frac1{2^3}-\frac1{2^n}\right)+\left(\frac1{2^4}-\frac1{2^n}\right)+\cdots+\left(\frac1{2^{n-2}}-\frac1{2^n}\right)+\left(\frac1{2^{n-1}}-\frac1{2^n}\right)

=\frac1{2^2}\left(1+\frac12+\frac1{2^2}+\cdots+\frac1{2^{n-3}}\right)-\frac{n-2}{2^n}

=\frac1{2^2}\cdot\frac{1-\frac1{2^{n-2}}}{1-\frac12}-\frac{n-2}{2^n}

=\frac12-\frac1{2^{n-1}}-\frac{n-2}{2^n}

=\frac12-\frac{n}{2^n}