Difference between revisions of "2023 AMC 8 Problems/Problem 16"
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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(" | + | label("P",(0.5,0.5)); |
− | label(" | + | label("Q",(1.5,0.5)); |
− | label(" | + | label("R",(2.5,0.5)); |
− | label(" | + | label("P",(3.5,0.5)); |
− | label(" | + | label("Q",(4.5,0.5)); |
− | label(" | + | label("Q",(0.5,1.5)); |
− | label(" | + | label("R",(1.5,1.5)); |
− | label(" | + | label("P",(2.5,1.5)); |
− | label(" | + | label("Q",(3.5,1.5)); |
− | label(" | + | label("R",(4.5,1.5)); |
− | label(" | + | label("R",(0.5,2.5)); |
− | label(" | + | label("P",(1.5,2.5)); |
− | label(" | + | label("Q",(2.5,2.5)); |
− | label(" | + | label("R",(3.5,2.5)); |
− | label(" | + | label("P",(4.5,2.5)); |
− | label(" | + | label("P",(0.5,3.5)); |
− | label(" | + | label("Q",(1.5,3.5)); |
− | label(" | + | label("R",(2.5,3.5)); |
− | label(" | + | label("P",(3.5,3.5)); |
− | label(" | + | label("Q",(4.5,3.5)); |
− | label(" | + | label("Q",(0.5,4.5)); |
− | label(" | + | label("R",(1.5,4.5)); |
− | label(" | + | label("P",(2.5,4.5)); |
− | label(" | + | label("Q",(3.5,4.5)); |
− | label(" | + | 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> | + | 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> | + | 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 05:19, 27 January 2023
Contents
Problem
The letters and
are entered into a
table according to the pattern shown below. How many
s,
s, and
s will appear in the completed table?
Solution 1
In our grid we can see there are
and
of the letters
and
’s respectively. We can see our pattern between each is
and
for the
and
’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
(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 of a letter. We find that it is
The rest of the grid with the P's and R's is symmetrical, so therefore, the answer is
Solution 3
Notice that rows and
are the same, for any
Additionally, rows
and
collectively contain the same number of
s,
s, and
s, because the letters are just substituted for one another. Therefore, the number of
s,
s, and
s in the first
rows is
. The first row has
s,
s, and
s, and the second row has
s,
s, and
s. Adding these up, we obtain
.
~mathboy100
Solution 4
From the full diagram below, the answer is
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
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using Cyclic Patterns)
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.