2023 AMC 12A Problems/Problem 20

Problem

Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below. $$ (Error compiling LaTeX. Unknown error_msg)\Longstack{ 1\\ 1\x 1\\ 1\x 2\x 1\\ 1\x 3\x 3\x 1\\ 1\x 4\x 6\x 4\x 1\\ 1\x 5\y 10\z 10\y 5\x 1\\ 1\x 6\y 15\z 20\z 15\y 6\x 1\\ \overline{0\x 1\x 2\x 3\x 4\x 5\x 6} } 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?

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)

$R(k) = 2^k - k$

$R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$

Now by definition from the question, the next row is always $:$ double the sum of last row (Each 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)

$2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$

Hence proven.

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

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
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2023 AMC 12A Problems/Problem 20

Problem

Solution 1

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