2023 AMC 12A Problems/Problem 20
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 1
First, let be the sum of the th row. Now, with some observation 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 each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1)
Which gives us
Hence, proven
Last, simply substitute , we get
Last digit of is ,
~lptoggled
Solution 2
Let the sum of the numbers in row be . Let each number in row be where .
Then From this we can establish:
Let
From this we have:
The problem requires us to find the last digit of . We can use modular arithmetic. ~luckuso
Solution 3 (Recursion)
Let the sum of the row be .
For each of the non-1 entries in the row, they are equal to the sum of the numbers diagonally above it in the row plus . Therefore all non-1 entries in the row appear twice in the sum of the non-1 entries in the row, the two s on each end of the row only appear once in the sum of the non-1 entries in the row. Additionally, additional s are placed at each end of the row. Hence,
, . By using the recursive formula,
is a geometric sequence by a ratio of
,
The unit digit of powers of is periodic by a cycle of digits: , , , . , the unit digit of is .
Therefore, the unit digit of is
Solution 4
Consider Pascal's triangle as the starting point. In the Pascal's triangle depicted below, the sum of the numbers in the nth row is . For the 2023rd row in the Pascal's triangle, the sum of numbers will be .
For this triangular array of integers in the problem, between the 3rd row and the 2022nd row, each 1 being added to the interior entry, which propagates to two numbers in the next row. Each 1 being added to a number in the Pascal's triangle eventually propagates to the 2023rd row twice. Therefore, the sum of numbers in the 2023rd row of this triangular array of integers will be the sum of numbers in the 2023rd row in the Pascal's triangle, plus two times all the 1's that have been added between the 3rd row and the 2022nd row, plus all the 1's of the 2021 interior entries in the 2023rd row.
Examine the following triangular array of 1's, there are 2020 rows between the 3rd and 2022nd rows, therefore the total number of 1's between the 3rd and 2022nd rows is .
The sum of numbers in the 2023rd row will be . For , the unit digit is the same as that of , which is 4. Therefore, the unit digit for the sum of numbers in the 2023rd row is .
~sqroot
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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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