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 
n
R(n) = 2^n - n
R(1) = 2^n - n = 2^1 - 1 = 1
R(k) = 2^k - k$$ (Error compiling LaTeX. Unknown error_msg)R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$Now by definition from the question, the next row is always: double the sum of last row (Each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (leftmost and rightmost are just 1)$ (Error compiling LaTeX. Unknown error_msg)2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$Hence proven.
Simply substitute$ (Error compiling LaTeX. Unknown error_msg)n = 2023
R(2023) = 2^2023 - 2023$
Last digit of 2^2023 is 8, 8-3 = \boxed{\textbf{(C)} 5}
~lptoggled
See also
| 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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