Difference between revisions of "2023 AMC 12A Problems/Problem 20"
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<math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math> | <math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math> | ||
| − | ==Solution | + | ==Solution== |
First, let <math>R(n)</math> 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>. | First, let <math>R(n)</math> 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>. | ||
Revision as of 00:38, 10 November 2023
Problem
Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
![[asy] size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$3$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$5$", (-0.5,-2)); label("$5$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$7$", (-1,-8/3)); label("$11$", (0,-8/3)); label("$7$", (1,-8/3)); label("$1$", (2,-8/3)); [/asy]](http://latex.artofproblemsolving.com/6/6/a/66a74e42d004740860a7d1c7d4eff043a98233ca.png)
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
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)
By definition from the question, the next row is always
Double the sum of last row (Imagine ach 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)
Which gives us
Hence proven
Last, 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. 
