Question

Rows $1,2,3,4,$ and $5$ of a triangular array of integers are shown below.

$\begin{array}{ccccccccc} &&&&1&&&&\\ &&&1&&1&&&\\ &&1&&3&&1&&\\ &1&&5&&5&&1&\\ 1&&7&&11&&7&&1 \end{array}$

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?

$1$

$3$

$5$

$7$

$9$

Solution:

Let $S_n$ be the sum of row $n.$ Each interior entry is $1$ more than the sum of the two entries above it, and summing over the row gives the recurrence $S_n=2S_{n-1}+(n-2).$

With $S_1=1,$ this solves to $S_n=2^n-n$ (check: $S_5=32-5=27=1+7+11+7+1$ ).

So $S_{2023}=2^{2023}-2023.$ Since powers of $2$ cycle with units digits $2,4,8,6$ and $2023\equiv 3\pmod 4,$ $2^{2023}$ ends in $8.$ Then $8-3=5$ gives units digit $5.$

Thus, the correct answer is C.

2023 AMC 12A Problem 20

Solution:

Problem 20 in Other Years