2023 AMC 8 Problems/Problem 16
Contents
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?
![[asy] /* Made by MRENTHUSIASM */ size(125); for (int y = 0; y<=5; ++y) { for (int x = 0; x<=5; ++x) { draw((x,0)--(x,6),mediumgrey); draw((0,y)--(6,y),mediumgrey); } } label("P",(0.5,0.5)); label("Q",(1.5,0.5)); label("R",(2.5,0.5)); label("P",(3.5,0.5)); label("Q",(4.5,0.5)); label("Q",(0.5,1.5)); label("R",(1.5,1.5)); label("P",(2.5,1.5)); label("Q",(3.5,1.5)); label("R",(4.5,1.5)); label("R",(0.5,2.5)); label("P",(1.5,2.5)); label("Q",(2.5,2.5)); label("R",(3.5,2.5)); label("P",(4.5,2.5)); label("P",(0.5,3.5)); label("Q",(1.5,3.5)); label("R",(2.5,3.5)); label("P",(3.5,3.5)); label("Q",(4.5,3.5)); label("Q",(0.5,4.5)); label("R",(1.5,4.5)); label("P",(2.5,4.5)); label("Q",(3.5,4.5)); label("R",(4.5,4.5)); label("$\vdots$",(0.5,5.5)); label("$\vdots$",(1.5,5.5)); label("$\vdots$",(2.5,5.5)); label("$\vdots$",(3.5,5.5)); label("$\vdots$",(4.5,5.5)); label("$\cdots$",(5.5,0.5)); label("$\cdots$",(5.5,1.5)); label("$\cdots$",(5.5,2.5)); label("$\cdots$",(5.5,3.5)); label("$\cdots$",(5.5,4.5)); label("$\cdot$",(5.3,5.3)); label("$\cdot$",(5.45,5.45)); label("$\cdot$",(5.6,5.6)); [/asy]](http://latex.artofproblemsolving.com/5/a/9/5a994064b982aa1cc082ae642d5bcbd1369fd477.png)
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
![[asy] /* Made by MRENTHUSIASM */ size(400); for (int y = 0; y<=20; ++y) { for (int x = 0; x<=20; ++x) { draw((x,0)--(x,20),mediumgrey); draw((0,y)--(20,y),mediumgrey); } } void drawDiagonal(string s, pair p) { while (p.x >= 0 && p.x < 20 && p.y >= 0 && p.y < 20) { label(s,p); p += (1,-1); } } drawDiagonal("P", (0.5,0.5)); drawDiagonal("Q", (0.5,1.5)); drawDiagonal("R", (0.5,2.5)); drawDiagonal("P", (0.5,3.5)); drawDiagonal("Q", (0.5,4.5)); drawDiagonal("R", (0.5,5.5)); drawDiagonal("P", (0.5,6.5)); drawDiagonal("Q", (0.5,7.5)); drawDiagonal("R", (0.5,8.5)); drawDiagonal("P", (0.5,9.5)); drawDiagonal("Q", (0.5,10.5)); drawDiagonal("R", (0.5,11.5)); drawDiagonal("P", (0.5,12.5)); drawDiagonal("Q", (0.5,13.5)); drawDiagonal("R", (0.5,14.5)); drawDiagonal("P", (0.5,15.5)); drawDiagonal("Q", (0.5,16.5)); drawDiagonal("R", (0.5,17.5)); drawDiagonal("P", (0.5,18.5)); drawDiagonal("Q", (0.5,19.5)); drawDiagonal("R", (1.5,19.5)); drawDiagonal("P", (2.5,19.5)); drawDiagonal("Q", (3.5,19.5)); drawDiagonal("R", (4.5,19.5)); drawDiagonal("P", (5.5,19.5)); drawDiagonal("Q", (6.5,19.5)); drawDiagonal("R", (7.5,19.5)); drawDiagonal("P", (8.5,19.5)); drawDiagonal("Q", (9.5,19.5)); drawDiagonal("R", (10.5,19.5)); drawDiagonal("P", (11.5,19.5)); drawDiagonal("Q", (12.5,19.5)); drawDiagonal("R", (13.5,19.5)); drawDiagonal("P", (14.5,19.5)); drawDiagonal("Q", (15.5,19.5)); drawDiagonal("R", (16.5,19.5)); drawDiagonal("P", (17.5,19.5)); drawDiagonal("Q", (18.5,19.5)); drawDiagonal("R", (19.5,19.5)); [/asy]](http://latex.artofproblemsolving.com/c/2/f/c2fcd5fccc751bb62a931a151eb88888d9beeb3e.png)
This solution is extremely time-consuming and error-prone. Please DO NOT try it in a real competition unless you have a large sheet of paper and you want to fail.
~MRENTHUSIASM
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
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. 
