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2023 AMC 8 Problems/Problem 16: Difference between revisions

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==Problem==
==Problem==


The letters <math>P, Q,</math> and <math>R</math> are entered into a <math>20\times20</math> table according to the pattern shown below. How many <math>P</math>s, <math>Q</math>s, and <math>R</math>s will appear in the completed table?  
The letters <math>\text{P}, \text{Q},</math> and <math>\text{R}</math> are entered into a <math>20\times20</math> table according to the pattern shown below. How many <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s will appear in the completed table?  
<asy>
<asy>
/* Made by MRENTHUSIASM */
/* Made by MRENTHUSIASM */
Line 13: Line 13:
}
}


label("$P$",(0.5,0.5));
label("P",(0.5,0.5));
label("$Q$",(1.5,0.5));
label("Q",(1.5,0.5));
label("$R$",(2.5,0.5));
label("R",(2.5,0.5));
label("$P$",(3.5,0.5));
label("P",(3.5,0.5));
label("$Q$",(4.5,0.5));
label("Q",(4.5,0.5));


label("$Q$",(0.5,1.5));
label("Q",(0.5,1.5));
label("$R$",(1.5,1.5));
label("R",(1.5,1.5));
label("$P$",(2.5,1.5));
label("P",(2.5,1.5));
label("$Q$",(3.5,1.5));
label("Q",(3.5,1.5));
label("$R$",(4.5,1.5));
label("R",(4.5,1.5));


label("$R$",(0.5,2.5));
label("R",(0.5,2.5));
label("$P$",(1.5,2.5));
label("P",(1.5,2.5));
label("$Q$",(2.5,2.5));
label("Q",(2.5,2.5));
label("$R$",(3.5,2.5));
label("R",(3.5,2.5));
label("$P$",(4.5,2.5));
label("P",(4.5,2.5));


label("$P$",(0.5,3.5));
label("P",(0.5,3.5));
label("$Q$",(1.5,3.5));
label("Q",(1.5,3.5));
label("$R$",(2.5,3.5));
label("R",(2.5,3.5));
label("$P$",(3.5,3.5));
label("P",(3.5,3.5));
label("$Q$",(4.5,3.5));
label("Q",(4.5,3.5));


label("$Q$",(0.5,4.5));
label("Q",(0.5,4.5));
label("$R$",(1.5,4.5));
label("R",(1.5,4.5));
label("$P$",(2.5,4.5));
label("P",(2.5,4.5));
label("$Q$",(3.5,4.5));
label("Q",(3.5,4.5));
label("$R$",(4.5,4.5));
label("R",(4.5,4.5));


label("$\vdots$",(0.5,5.5));
label("$\vdots$",(0.5,5.5));
Line 71: Line 71:
== Solution 1 ==
== Solution 1 ==


In our <math>5 \times 5</math> grid we can see there are <math>8,9</math> and <math>8</math> of the letters <math>P, Q,</math> and <math>R</math>’s respectively. We can see our pattern between each is <math>x, x+1,</math> and <math>x</math> for the <math>P, Q,</math> and <math>R</math>’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 <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math>
In our <math>5 \times 5</math> grid we can see there are <math>8,9</math> and <math>8</math> of the letters <math>\text{P}, \text{Q},</math> and <math>\text{R}</math>’s respectively. We can see our pattern between each is <math>x, x+1,</math> and <math>x</math> for the <math>\text{P}, \text{Q},</math> and <math>\text{R}</math>’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 <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math>


(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.)
(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.)
Line 79: Line 79:
== Solution 2 ==
== Solution 2 ==


We think about which letter is in the diagonal with <math>20</math> of a letter. We find that it is <math>2(2 + 5 + 8 + 11 + 14 + 17) + 20 = 134</math>. The rest of the grid with the P's and R's is symmetrical, so therefore, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>.
We think about which letter is in the diagonal with <math>20</math> of a letter. We find that it is <math>2(2 + 5 + 8 + 11 + 14 + 17) + 20 = 134.</math> The rest of the grid with the P's and R's is symmetrical, so therefore, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math>


~[[User:ILoveMath31415926535|ILoveMath31415926535]]
~[[User:ILoveMath31415926535|ILoveMath31415926535]]
Line 85: Line 85:
== Solution 3 ==
== 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{\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>.


~mathboy100
~mathboy100
Line 92: Line 92:


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>
<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>
<b>This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.</b>
<b>This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.</b>


Revision as of 04:19, 27 January 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 */ 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 (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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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