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

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We know that all the squares with side lengths that are multiples of 3 will have an equal number of Ps Qs, and Rs, so we don't worry too much about the 3x3. From the other ones, we can easily see that they are all <math>x</math>, <math>x+1</math>, and <math>x</math>. Using this, we find the only option that provides this format, which is  <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>
 
We know that all the squares with side lengths that are multiples of 3 will have an equal number of Ps Qs, and Rs, so we don't worry too much about the 3x3. From the other ones, we can easily see that they are all <math>x</math>, <math>x+1</math>, and <math>x</math>. Using this, we find the only option that provides this format, which is  <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>
 +
 +
~CoOlPoTaToEs
  
 
==Video Solution by Math-X (Let's first Understand the question)==
 
==Video Solution by Math-X (Let's first Understand the question)==

Revision as of 23:23, 22 January 2024

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 (LOGIC)

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

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 (Brute-Force)

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

Solution 5

This solution refers to the full diagram in Solution 4.

Note the $\text{Q}$ diagonals are symmetric. The $\text{R}$ and $\text{P}$ diagonals are not symmetric, but are reflections of each other about the $\text{Q}$ diagonals:

  • The upper $\text{Q}$ diagonal of length $2$ is surrounded by a $\text{P}$ diagonal of length $3$ and an $\text{R}$ diagonal of length $1.$
  • The lower $\text{Q}$ diagonal of length $2$ is surrounded by a $\text{P}$ diagonal of length $1$ and an $\text{R}$ diagonal of length $3.$

When looking at a pair of $Q$ diagonals of the same length $x,$ there is a total of $2x$ $\text{R}$s and $\text{P}$s next to these $2$ diagonals.

The main diagonal of $20$ $\text{Q}$s has $19$ $\text{P}$s and $19$ $\text{R}$s next to it. Thus, the total is $x+1$ $\text{Q}$s, $x$ $\text{P}$s, $x$ $\text{R}$s. Therefore, the answer is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$

~ERMSCoach

Solution 6 (Modular Arithmetic)

Note that the a different letter is always to either the left and right to one letter (there will never be two of the same letter in a row or column).

It follows that if a letter is in a horizontal position $k$, then that same letter will appear in position $k+3m$, for a positive integer $m$. In other words, all positions congruent to $k$ modulo $3$ will have the same letter as $p$.

Since $p$ is in position $1$, $p$ will be in every position congruent to $1 \pmod 3$. There are $7$ numbers less than or equal to $20$ that satisfy this restraint. There are also $7$ numbers less than or equal to $2$ that are congruent to $2 \pmod 3$, but only $6$ that are multiples of $3$.

In $p$'s case, it will appear $7$ times in row one, only $6$ in row $2$ (as its first appearence is in position $3$), and $7$ in row $3$. So in the first $3$ rows, $P$ appears $20$ times.

Therefore, in the first $18$ rows, $P$ appears $20\cdot 6 = 120$ times. Row $19$ looks identical to row $1$, as $19\equiv 1\pmod 3$, so $P$ appears in row $19$ $7$ times. It follows that $P$ appears in row $20$ $6$ times. There are $134 P$s.

Counting $Q$s is nearly identical, but $Q$ begins in position $2$. In the first $18$ rows, there are an identical amount of $Q$s too, namely $6(7+7+6) = 120$. However, by a similar argument to $P$, $Q$ appears $7+7$ times in the last two rows; because row $20$ is the same as row $2$, $Q$ appears in position $1$, and thus $7$ times.

Therefore, there are $120+14 = 134$ $Q$s and $133$ $P$. We could go through a similar argument for $R$, or note that the only answer choice with these two options is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$

-Benedict T (countmath1)

Solution 7 (Answer Choices, Fast)

We can first observe that $P$ shows on diagonals increasing or decreasing by $3.$

It starts at $1,4,7,9...$ and increases in the form $3x-2$. Using our answer choices, $(B)$ and $(C)$ are the only fits.

$Q$ is like this as well, increasing $2,5,8,11....$ This means $Q$ has to be in the form of $3x-1.$ Testing this out leads us with $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.$ ~andyluo


Solution 8 (Patterns)

We can start by writing down the Ps, Qs, and Rs for 5x5, 4x4, 3x3, and 2x2 squares:

  • 5x5: 8,9,8
  • 4x4: 6,5,5
  • 3x3: 3,3,3
  • 2x2: 1,2,1

We know that all the squares with side lengths that are multiples of 3 will have an equal number of Ps Qs, and Rs, so we don't worry too much about the 3x3. From the other ones, we can easily see that they are all $x$, $x+1$, and $x$. Using this, we find the only option that provides this format, which is $\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}$

~CoOlPoTaToEs

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=Xxe3CXtrXQYGSYik&t=3081

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/3DTwjLe0Pw0

~Education, the Study of Everything

Video Solution (Animated)

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

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=1845

Video Solution by WhyMath

https://youtu.be/iMhqlz0-ce0

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=jtHe6yLBz-4&t=20s

~harungurcan

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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