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2023 AMC 8 Problems/Problem 16

Revision as of 09:07, 23 January 2024 by Nivaar (talk | contribs) (→‎Solution 1 (Logic))

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/Finding Patterns)

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 fast" 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 $P$s, $Q$s, and $R$s for $5\times5, 4\times4, 3\times3,$ and $2\times2$ squares:

  • $5\times5: \ 8,9,8$
  • $4\times4: \ 6,5,5$
  • $3\times3: \ 3,3,3$
  • $2\times2: \ 1,2,1$

We know that all the squares with side lengths that are multiples of $3$ will have an equal number of $P$s, $Q$s, and $R$s, so we don't worry too much about the $3\times3.$ 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

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