2023 AMC 8 Problems/Problem 16
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Modular Arithmetic)
- 8 Animated Video Solution
- 9 Video Solution by OmegaLearn (Using Cyclic Patterns)
- 10 Video Solution by Magic Square
- 11 Video Solution by Interstigation
- 12 Video Solution by WhyMath
- 13 Video Solution (CREATIVE THINKING!!!)
- 14 Video Solution by harungurcan
- 15 See Also
Problem
The letters and are entered into a table according to the pattern shown below. How many s, s, and s will appear in the completed table?
Solution 1
In our grid we can see there are and of the letters and ’s respectively. We can see our pattern between each is and for the and ’s respectively. This such pattern will follow in our bigger example, so we can see that the only answer choice which satisfies this condition is
(Note: you could also "cheese" this problem by listing out all of the letters horizontally in a single line and looking at the repeating pattern.)
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
Solution 2
We think about which letter is in the diagonal with of a letter. We find that it is The rest of the grid with the 's and 's is symmetric. Therefore, the answer is
Solution 3
Notice that rows and are the same, for any Additionally, rows and collectively contain the same number of s, s, and s, because the letters are just substituted for one another. Therefore, the number of s, s, and s in the first rows is . The first row has s, s, and s, and the second row has s, s, and s. Adding these up, we obtain .
~mathboy100
Solution 4
From the full diagram below, the answer is This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.
~MRENTHUSIASM
Solution 5
This solution refers to the full diagram in Solution 4.
Note the diagonals are symmetric. The and diagonals are not symmetric, but are reflections of each other about the diagonals:
- The upper diagonal of length is surrounded by a diagonal of length and an diagonal of length
- The lower diagonal of length is surrounded by a diagonal of length and an diagonal of length
When looking at a pair of diagonals of the same length there is a total of s and s next to these diagonals.
The main diagonal of s has s and s next to it. Thus, the total is s, s, s. Therefore, the answer is
~ERMSCoach
Solution 6 (Modular Arithmetic)
Note that the a different letter is always to either the left and right to one letter (there will never be two of the same letter in a row or column).
It follows that if a letter is in a horizontal position , then that same letter will appear in position , for a positive integer . In other words, all positions congruent to modulo will have the same letter as .
Since is in position , will be in every position congruent to . There are numbers less than or equal to that satisfy this restraint. There are also numbers less than or equal to that are congruent to , but only that are multiples of .
In 's case, it will appear times in row one, only in row (as its first appearence is in position ), and in row . So in the first rows, appears times.
Therefore, in the first rows, appears times. Row looks identical to row , as , so appears in row times. It follows that appears in row times. There are s.
Counting s is nearly identical, but begins in position . In the first rows, there are an identical amount of s too, namely . However, by a similar argument to , appears times in the last two rows; because row is the same as row , appears in position , and thus times.
Therefore, there are s and . We could go through a similar argument for , or note that the only answer choice with these two options is
-Benedict T (countmath1)
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using Cyclic Patterns)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3990
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=1385
Video Solution by WhyMath
~savannahsolver
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution by harungurcan
https://www.youtube.com/watch?v=jtHe6yLBz-4&t=20s
~harungurcan
See Also
2023 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.