2023 AMC 12A Problems/Problem 20

Problem

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

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
All AMC 12 Problems and Solutions

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2023 AMC 12A Problems/Problem 20

Problem

Solution 1

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