2023 AMC 8 Problems/Problem 16

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Problem

The letters $\text{P}, \text{Q},$ and $\text{R}$ are entered into a $20\times20$ table according to the pattern shown below. How many $\text{P}$s, $\text{Q}$s, and $\text{R}$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 $\text{P}, \text{Q},$ and $\text{R}$’s respectively. We can see our pattern between each is $x, x+1,$ and $x$ for the $\text{P}, \text{Q},$ and $\text{R}$’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 $\text{P}$s, $\text{Q}$s, and $\text{R}$s, because the letters are just substituted for one another. Therefore, the number of $\text{P}$s, $\text{Q}$s, and $\text{R}$s in the first $18$ rows is $120$. The first row has $7$ $\text{P}$s, $7$ $\text{Q}$s, and $6$ $\text{R}$s, and the second row has $6$ $\text{P}$s, $7$ $\text{Q}$s, and $7$ $\text{R}$s. Adding these up, we obtain $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}$.

~mathboy100

Solution 4

From the full diagram below, the answer is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$ [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] 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

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 (ProblemsAnswer KeyResources)
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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