Difference between revisions of "2023 AMC 8 Problems/Problem 16"
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We think about which letter is in the diagonal with <math>20</math> of a letter. We find that it is <math>2(2 + 5 + 8 + 11 + 14 + 17) + 20 = 134</math>. The rest of the grid with the P's and R's is symmetrical, so therefore, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>. | We think about which letter is in the diagonal with <math>20</math> of a letter. We find that it is <math>2(2 + 5 + 8 + 11 + 14 + 17) + 20 = 134</math>. The rest of the grid with the P's and R's is symmetrical, so therefore, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>. | ||
| − | + | ||
| + | ~[[User:ILoveMath31415926535|ILoveMath31415926535]] | ||
== Solution 3 == | == Solution 3 == | ||
Revision as of 03:49, 27 January 2023
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/1/8/7/1871978bb8c9bd31795b1df18384f105716378a6.png)
Solution 1
In our 
grid we can see there are 
, 
and of the letters P, Q and R’s respectively. We can see our pattern between each is 
, 
, for the P, Q and R’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 P's and R's is symmetrical, so 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 
, 
, and 
, and the second row has 
, , and 
. Adding these up, we obtain .
~mathboy100
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. 
