Difference between revisions of "2023 AMC 12A Problems/Problem 20"
(→Solution 1) |
m (→Solution 1) |
||
| Line 22: | Line 22: | ||
Simply substitute <math>n = 2023</math>, we get <math>R(2023) = 2^2023 - 2023</math> | Simply substitute <math>n = 2023</math>, we get <math>R(2023) = 2^2023 - 2023</math> | ||
| − | Last digit of <math>2^{2023}</math> is <math>8</math>, <math>8-3 = \boxed{\textbf{(C)} 5}</math> | + | Last digit of <math>2^{2023}</math> is <math>8</math>, <math>8-3 = \boxed{\textbf{(C) } 5}</math> |
~lptoggled | ~lptoggled | ||
Revision as of 21:45, 9 November 2023
Problem
Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
Solution 1
First, let 
be the sum of the 
th row. Now, with some observations and math instinct, we can guess that 
.
now we try to prove it by induction,
(works for base case)
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)
Hence proven.
Simply substitute 
, we get 
Last digit of 
is 
, 
~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. 
