Difference between revisions of "2023 AMC 12A Problems/Problem 20"
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<math>2^{2023} - 2023 \equiv 8 - 3\pmod{10}</math> | <math>2^{2023} - 2023 \equiv 8 - 3\pmod{10}</math> | ||
− | <math>\equiv 5 \pmod{10}</math> | + | <math>\equiv \boxed{\textbf{(C) } 5} \pmod{10}</math> |
~cyantist | ~cyantist |
Revision as of 02:25, 10 November 2023
Contents
Problem
Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
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 2
Let the sum of the numbers in row be
. Let each number in row
be
where
.
Then
From this we can establish 2 equations:
Let
The problem requires us to find the last digit of . We can use modular arithmetic.
~cyantist
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)
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 (you can work it backwards without induction & guessing)
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.