2023 AMC 12A Problem 20

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

Let

S_n

be the sum of row

n;

each interior entry adds two copies of the row above plus a

1

This yields

S_n=2S_{n-1}+(n-2),

whose closed form is

S_n=2^n-n

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.

Problem 20 in other years