Difference between revisions of "2023 AMC 12A Problems/Problem 20"
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| − | ( | + | ==Problem== |
| + | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below. | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | First, let <math>R(n) be the sum of the </math>n<math>th row. Now, with some observations and math instinct, we can guess that </math>R(n) = 2^n - n<math> | ||
| + | |||
| + | now we try to prove it by induction, | ||
| + | |||
| + | </math>R(1) = 2^n - n = 2^1 - 1 = 1<math> (works for base case!) | ||
| + | |||
| + | </math>R(k) = 2^k - k<math> | ||
| + | |||
| + | </math>R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1<math> | ||
| + | |||
| + | 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) | ||
| + | |||
| + | </math>2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1<math> | ||
| + | |||
| + | Hence proven. | ||
| + | |||
| + | Simply substitute </math>n = 2023<math>, we get </math>R(2023) = 2^2023 - 2023$ | ||
| + | |||
| + | Last digit of 2^2023 is 8, 8-3 = \boxed{\textbf{(C)} 5} | ||
| + | |||
| + | ~lptoggled | ||
| + | |||
| + | ==See also== | ||
| + | |||
| + | {{AMC12 box|year=2023|ab=A|num-b=19|num-a=21}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 21:42, 9 November 2023
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
