Difference between revisions of "2023 AMC 8 Problems/Problem 16"
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<math>\textbf{(E)}~134\text{ Ps, }133\text{ Qs, }133\text{ Rs}</math> | <math>\textbf{(E)}~134\text{ Ps, }133\text{ Qs, }133\text{ Rs}</math> | ||
− | == Solution 1 == | + | == Solution 1 (Logic/Finding Patterns)== |
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> | 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 brute force/listing out all of the letters horizontally in a single line and looking at the repeating pattern. Refer to solution 4) |
− | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | + | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, Nivaar |
== Solution 2 == | == Solution 2 == | ||
Line 162: | Line 162: | ||
drawDiagonal("R", (19.5,19.5)); | drawDiagonal("R", (19.5,19.5)); | ||
</asy> | </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>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> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Line 198: | Line 198: | ||
-Benedict T (countmath1) | -Benedict T (countmath1) | ||
+ | |||
+ | == Solution 7 (Answer Choices, Fast)== | ||
+ | |||
+ | We can first observe that <math>P</math> shows on diagonals increasing or decreasing by <math>3.</math> | ||
+ | |||
+ | It starts at <math>1,4,7,9... </math> and increases in the form <math>3x-2</math>. Using our answer choices, <math>(B)</math> and <math>(C)</math> are the only fits. | ||
+ | |||
+ | <math>Q</math> is like this as well, increasing <math>2,5,8,11....</math> This means <math>Q</math> has to be in the form of <math>3x-1.</math> Testing this out leads us with <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math> | ||
+ | |||
+ | ~andyluo | ||
+ | |||
+ | ==Solution 8 (Patterns)== | ||
+ | NOTE THAT THIS SOLUTION IS VERY SIMILAR TO SOLUTION 1. | ||
+ | |||
+ | We can start by writing down the <math>P</math>s, <math>Q</math>s, and <math>R</math>s for <math>5\times5, 4\times4, 3\times3,</math> and <math>2\times2</math> squares: | ||
+ | |||
+ | * <math>5\times5: \ 8,9,8</math> | ||
+ | * <math>4\times4: \ 6,5,5</math> | ||
+ | * <math>3\times3: \ 3,3,3</math> | ||
+ | * <math>2\times2: \ 1,2,1</math> | ||
+ | |||
+ | We know that all the squares with side lengths that are multiples of <math>3</math> will have an equal number of <math>P</math>s, <math>Q</math>s, and <math>R</math>s, so we don't worry too much about the <math>3\times3.</math> From the other ones, we can easily see that they are all <math>x, 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 (Solve under 60 seconds!!!)== | ||
+ | https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=731 | ||
+ | |||
+ | ~hsnacademy | ||
==Video Solution by Math-X (Let's first Understand the question)== | ==Video Solution by Math-X (Let's first Understand the question)== |
Latest revision as of 02:07, 21 August 2024
Contents
- 1 Problem
- 2 Solution 1 (Logic/Finding Patterns)
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Brute-Force)
- 6 Solution 5
- 7 Solution 6 (Modular Arithmetic)
- 8 Solution 7 (Answer Choices, Fast)
- 9 Solution 8 (Patterns)
- 10 Video Solution (Solve under 60 seconds!!!)
- 11 Video Solution by Math-X (Let's first Understand the question)
- 12 Video Solution (CREATIVE THINKING!!!)
- 13 Video Solution (Animated)
- 14 Video Solution by OmegaLearn (Using Cyclic Patterns)
- 15 Video Solution by Magic Square
- 16 Video Solution by Interstigation
- 17 Video Solution by WhyMath
- 18 Video Solution by harungurcan
- 19 See Also
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 (Logic/Finding Patterns)
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 brute force/listing out all of the letters horizontally in a single line and looking at the repeating pattern. Refer to solution 4)
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, Nivaar
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
's and
's is symmetric. 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 (Brute-Force)
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
Solution 5
This solution refers to the full diagram in Solution 4.
Note the diagonals are symmetric. The
and
diagonals are not symmetric, but are reflections of each other about the
diagonals:
- The upper
diagonal of length
is surrounded by a
diagonal of length
and an
diagonal of length
- The lower
diagonal of length
is surrounded by a
diagonal of length
and an
diagonal of length
When looking at a pair of diagonals of the same length
there is a total of
s and
s next to these
diagonals.
The main diagonal of
s has
s and
s next to it. Thus, the total is
s,
s,
s. Therefore, the answer is
~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 , then that same letter will appear in position
, for a positive integer
. In other words, all positions congruent to
modulo
will have the same letter as
.
Since is in position
,
will be in every position congruent to
. There are
numbers less than or equal to
that satisfy this restraint. There are also
numbers less than or equal to
that are congruent to
, but only
that are multiples of
.
In 's case, it will appear
times in row one, only
in row
(as its first appearence is in position
), and
in row
. So in the first
rows,
appears
times.
Therefore, in the first rows,
appears
times. Row
looks identical to row
, as
, so
appears in row
times. It follows that
appears in row
times. There are
s.
Counting s is nearly identical, but
begins in position
. In the first
rows, there are an identical amount of
s too, namely
. However, by a similar argument to
,
appears
times in the last two rows; because row
is the same as row
,
appears in position
, and thus
times.
Therefore, there are
s and
. We could go through a similar argument for
, or note that the only answer choice with these two options is
-Benedict T (countmath1)
Solution 7 (Answer Choices, Fast)
We can first observe that shows on diagonals increasing or decreasing by
It starts at and increases in the form
. Using our answer choices,
and
are the only fits.
is like this as well, increasing
This means
has to be in the form of
Testing this out leads us with
~andyluo
Solution 8 (Patterns)
NOTE THAT THIS SOLUTION IS VERY SIMILAR TO SOLUTION 1.
We can start by writing down the s,
s, and
s for
and
squares:
We know that all the squares with side lengths that are multiples of will have an equal number of
s,
s, and
s, so we don't worry too much about the
From the other ones, we can easily see that they are all
and
Using this, we find the only option that provides this format, which is
~CoOlPoTaToEs
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=KvvABit-3-ZtX7Qa&t=731
~hsnacademy
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!!!)
~Education, the Study of Everything
Video Solution (Animated)
~Star League (https://starleague.us)
Video Solution by OmegaLearn (Using Cyclic Patterns)
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
~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.