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

Revision as of 15:10, 25 January 2023

Problem

The letters P, Q, and R are entered into a $20\times20$ table according to the pattern shown below. How many Ps, Qs, and Rs will appear in the completed table?

$%%Table made by Technodoggo \begin{array}[b]{|c|c|c|c|c|c} \vdots &\vdots&\vdots&\vdots&\vdots&\iddots\\\hline Q&R&P&Q&R&\cdots\\\hline P&Q&R&P&Q&\cdots\\\hline R&P&Q&R&P&\cdots\\\hline Q&R&P&Q&R&\cdots\\\hline P&Q&R&P&Q&\cdots\\\hline \end{array}$

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

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

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

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

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

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 $x$ , $x+1$ , $x$ 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 $\boxed{\text{(C)}\hspace{0.1 in} 133, 134, 133}$

(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 $\boxed{\text{(C)}\hspace{0.1 in} 133, 134, 133}$

Solution by 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 $\boxed{\text{(C)}\hspace{0.1 in} 133, 134, 133}$ .

~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

@@ Line 41: / Line 41: @@
 Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]]
+== Solution 3 ==
+Notice that rows <math>x</math> and <math>x+3</math> are the same, for any <math>1 \leq x \leq 17</math>. Additionally, rows <math>1</math>, <math>2</math>, and <math>3</math> collectively contain the same number of <math>P</math>s, <math>Q</math>s, and <math>R</math>s, because the letters are just substituted for one another. Therefore, the number of <math>P</math>s, <math>Q</math>s, and <math>R</math>s in the first <math>18</math> rows is <math>120</math>. The first row has <math>7P</math>, <math>7Q</math>, and <math>6R</math>, and the second row has <math>6P</math>, <math>7Q</math>, and <math>7R</math>. Adding these up, we obtain <math>\boxed{\text{(C)}\hspace{0.1 in} 133, 134, 133}</math>.
+~mathboy100
 ==Animated Video Solution==

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
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