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Difference between revisions of "2023 AMC 8 Problems/Problem 16"

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== Solution 1 ==
 
== Solution 1 ==
  
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In our <math>5 \times 5</math> grid we can see there are <math>8</math>, <math>9</math> and <math>8</math> of the letters P, Q and R’s respectively. We can see our pattern between each is <math>x</math>, <math>x+1</math>, <math>x</math> 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 <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>
+
In our <math>5 \times 5</math> grid we can see there are <math>8,9</math> and <math>8</math> of the letters <math>P, Q,</math> and <math>R</math>’s respectively. We can see our pattern between each is <math>x, x+1,</math> and <math>x</math> for the <math>P, Q,</math> and <math>R</math>’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 <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math>
  
 
(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.)
 
(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.)

Revision as of 03:50, 27 January 2023

Problem

The letters $P, Q,$ and $R$ are entered into a $20\times20$ table according to the pattern shown below. How many $P$s, $Q$s, and $R$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] $\textbf{(A)}~132\text{ Ps, }134\text{ Qs, }134\text{ Rs}$

$\textbf{(B)}~133\text{ Ps, }133\text{ Qs, }134\text{ Rs}$

$\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}$

$\textbf{(D)}~134\text{ Ps, }132\text{ Qs, }134\text{ Rs}$

$\textbf{(E)}~134\text{ Ps, }133\text{ Qs, }133\text{ Rs}$

Solution 1

In our $5 \times 5$ grid we can see there are $8,9$ and $8$ of the letters $P, Q,$ and $R$’s respectively. We can see our pattern between each is $x, x+1,$ and $x$ 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 $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$

(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 $20$ of a letter. We find that it is $2(2 + 5 + 8 + 11 + 14 + 17) + 20 = 134$. The rest of the grid with the P's and R's is symmetrical, so therefore, the answer is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}$.

~ILoveMath31415926535

Solution 3

Notice that rows $x$ and $x+3$ are the same, for any $1 \leq x \leq 17$. Additionally, rows $1$, $2$, and $3$ collectively contain the same number of $P$s, $Q$s, and $R$s, because the letters are just substituted for one another. Therefore, the number of $P$s, $Q$s, and $R$s in the first $18$ rows is $120$. The first row has $7P$, $7Q$, and $6R$, and the second row has $6P$, $7Q$, and $7R$. Adding these up, we obtain $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}$. ~mathboy100

Animated Video Solution

https://youtu.be/1tnMR0lNEFY

~Star League (https://starleague.us)

Video Solution by OmegaLearn (Using Cyclic Patterns)

https://youtu.be/83FnFhe4QgQ

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

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