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

(Using symmetry)
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== Solution 3 ==
 
== Solution 3 ==
 +
Note the Q diagonals are symmetrical.  The R and P diagonals are not symmetrical, but are reflections of each other- the Q diagonals of length 2 are surrounded by a P of length 1 and an R of length 3, and a P of length 3 and an R of length 1.
 +
When looking at a pair of Q diagonals of the same length x, there is a total of 2x Rs and Ps next to these 2 diagonals.
 +
The main diagonal of 20Q has 19 P and 19R next to it.  Thus the total is x Q, x+1 P, x+1 R, which is 133, 134, 133 PQR.
 +
 +
~ERMSCoach
 +
 +
== Solution 4 ==
  
 
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, 2,</math> and <math>3</math> collectively contain the same number of <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s, because the letters are just substituted for one another. Therefore, the number of <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s in the first <math>18</math> rows is <math>120</math>. The first row has <math>7</math> <math>\text{P}</math>s, <math>7</math> <math>\text{Q}</math>s, and <math>6</math> <math>\text{R}</math>s, and the second row has <math>6</math> <math>\text{P}</math>s, <math>7</math> <math>\text{Q}</math>s, and <math>7</math> <math>\text{R}</math>s. Adding these up, we obtain <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>.
 
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, 2,</math> and <math>3</math> collectively contain the same number of <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s, because the letters are just substituted for one another. Therefore, the number of <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s in the first <math>18</math> rows is <math>120</math>. The first row has <math>7</math> <math>\text{P}</math>s, <math>7</math> <math>\text{Q}</math>s, and <math>6</math> <math>\text{R}</math>s, and the second row has <math>6</math> <math>\text{P}</math>s, <math>7</math> <math>\text{Q}</math>s, and <math>7</math> <math>\text{R}</math>s. Adding these up, we obtain <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>.
Line 92: Line 99:
 
~mathboy100
 
~mathboy100
  
−
== Solution 4 ==
+
== Solution 5 ==
  
 
From the full diagram below, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math>
 
From the full diagram below, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math>
Line 165: Line 172:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
−
 
−
== Solution 5 ==
 
−
Note the Q diagonals are symmetrical.  The R and P diagonals are not symmetrical, but are reflections of each other- the Q diagonals of length 2 are surrounded by a P of length 1 and an R of length 3, and a P of length 3 and an R of length 1.
 
−
When looking at a pair of Q diagonals of the same length x, there is a total of 2x Rs and Ps next to these 2 diagonals.
 
−
The main diagonal of 20Q has 19 P and 19R next to it.  Thus the total is x Q, x+1 P, x+1 R, which is 133, 134, 133 PQR.
 
−
 
−
~ERMSCoach
 
  
 
==Animated Video Solution==
 
==Animated Video Solution==

Revision as of 00:22, 3 February 2023

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, Edited by Kante314 */ usepackage("mathdots"); size(5cm); draw((0,0)--(6,0),linewidth(1.5)+mediumgray); draw((0,1)--(6,1),linewidth(1.5)+mediumgray); draw((0,2)--(6,2),linewidth(1.5)+mediumgray); draw((0,3)--(6,3),linewidth(1.5)+mediumgray); draw((0,4)--(6,4),linewidth(1.5)+mediumgray); draw((0,5)--(6,5),linewidth(1.5)+mediumgray); draw((0,0)--(0,6),linewidth(1.5)+mediumgray); draw((1,0)--(1,6),linewidth(1.5)+mediumgray); draw((2,0)--(2,6),linewidth(1.5)+mediumgray); draw((3,0)--(3,6),linewidth(1.5)+mediumgray); draw((4,0)--(4,6),linewidth(1.5)+mediumgray); draw((5,0)--(5,6),linewidth(1.5)+mediumgray); label(scale(.9)*"\textsf{P}", (.5,.5)); label(scale(.9)*"\textsf{Q}", (.5,1.5)); label(scale(.9)*"\textsf{R}", (.5,2.5)); label(scale(.9)*"\textsf{P}", (.5,3.5)); label(scale(.9)*"\textsf{Q}", (.5,4.5)); label("$\vdots$", (.5,5.6)); label(scale(.9)*"\textsf{Q}", (1.5,.5)); label(scale(.9)*"\textsf{R}", (1.5,1.5)); label(scale(.9)*"\textsf{P}", (1.5,2.5)); label(scale(.9)*"\textsf{Q}", (1.5,3.5)); label(scale(.9)*"\textsf{R}", (1.5,4.5)); label("$\vdots$", (1.5,5.6)); label(scale(.9)*"\textsf{R}", (2.5,.5)); label(scale(.9)*"\textsf{P}", (2.5,1.5)); label(scale(.9)*"\textsf{Q}", (2.5,2.5)); label(scale(.9)*"\textsf{R}", (2.5,3.5)); label(scale(.9)*"\textsf{P}", (2.5,4.5)); label("$\vdots$", (2.5,5.6)); label(scale(.9)*"\textsf{P}", (3.5,.5)); label(scale(.9)*"\textsf{Q}", (3.5,1.5)); label(scale(.9)*"\textsf{R}", (3.5,2.5)); label(scale(.9)*"\textsf{P}", (3.5,3.5)); label(scale(.9)*"\textsf{Q}", (3.5,4.5)); label("$\vdots$", (3.5,5.6)); label(scale(.9)*"\textsf{Q}", (4.5,.5)); label(scale(.9)*"\textsf{R}", (4.5,1.5)); label(scale(.9)*"\textsf{P}", (4.5,2.5)); label(scale(.9)*"\textsf{Q}", (4.5,3.5)); label(scale(.9)*"\textsf{R}", (4.5,4.5)); label("$\vdots$", (4.5,5.6)); label(scale(.9)*"$\dots$", (5.5,.5)); label(scale(.9)*"$\dots$", (5.5,1.5)); label(scale(.9)*"$\dots$", (5.5,2.5)); label(scale(.9)*"$\dots$", (5.5,3.5)); label(scale(.9)*"$\dots$", (5.5,4.5)); label(scale(.9)*"$\iddots$", (5.5,5.6)); [/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 $\text{P}$'s and $\text{R}$'s is symmetric. Therefore, the answer is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$

~ILoveMath31415926535

Solution 3

Note the Q diagonals are symmetrical. The R and P diagonals are not symmetrical, but are reflections of each other- the Q diagonals of length 2 are surrounded by a P of length 1 and an R of length 3, and a P of length 3 and an R of length 1. When looking at a pair of Q diagonals of the same length x, there is a total of 2x Rs and Ps next to these 2 diagonals. The main diagonal of 20Q has 19 P and 19R next to it. Thus the total is x Q, x+1 P, x+1 R, which is 133, 134, 133 PQR.

~ERMSCoach

Solution 4

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 5

From the full diagram below, the answer is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$ [asy] /* Made by MRENTHUSIASM, Edited by Kante314 */ usepackage("mathdots"); size(16.666cm); for (int y = 0; y<=20; ++y) { for (int x = 0; x<=20; ++x) { draw((x,0)--(x,20),linewidth(1.5)+mediumgray); draw((0,y)--(20,y),linewidth(1.5)+mediumgray); } } void drawDiagonal(string s, pair p) { while (p.x >= 0 && p.x < 20 && p.y >= 0 && p.y < 20) { label(scale(.9)*("\textsf{" + 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 (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. AMC logo.png

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