2023 AMC 8 Problems/Problem 16
Contents
[hide]- 1 Problem
- 2 Solution 1 (Finding a Pattern)
- 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 (Intuitive, Fastest)
- 10 Solution 9 (Think and Label)
- 11 Video Solution 1 by Math-X
- 12 Video Solution 2 by CoolMathProblems
- 13 Video Solution 3 by hnsacademy
- 14 Video Solution 4
- 15 Video Solution 5
- 16 Video Solution 6 by OmegaLearn (Using Cyclic Patterns)
- 17 Video Solution 7 by Magic Square
- 18 Video Solution 8 by Interstigation
- 19 Video Solution 9 by WhyMath
- 20 Video Solution 10
- 21 Video Solution 11 by Dr. David
- 22 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 (Finding a Pattern)
In our grid, there are
and
of the letters
and
, respectively, and in a
grid, there are
and
of the letters
and
, respectively. We see that in both grids, there are
and
of the
and
, respectively. This is because in any
grid with
, there are
and
of the
and
, respectively. We can see that the only answer choice which satisfies this condition is
~CoOlPoTaToEs, apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, Nivaar
Solution 2
Since and
is in the 2nd diagonal, it is also in the 20th diagonal, and so we find that there are
. Since all the
's and
's are symmetric, 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, the answer is
This solution is extremely time-consuming and error-prone. 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)
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 appearance 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 (Intuitive, Fastest)
When we find the letters at the corners, we see that there are 2 other corners with Q's and 1 with R. There are 2 corners with Q and 1 with P and R. Therefore, it makes sense that there should be more Q's than other letters. Only has more Q's than other letters.
Only do it this way if there are 30 seconds left on the clock, as it may not always work!
Solution 9 (Think and Label)
Since there are letters, there must be a multiple of 3 for the height of the array for the amount of letters to be the same. However, there is one less (
) so there will be one less letter in each column compared to the other two. Looking at the diagram, the first column starts with a
then a
. Since it will miss the third letter,
, we write this as
. We do this process for the remaining 20 columns, and applying the same logic if the first three are
,
and
, there will be one less
meaning that there will be an extra
. The only answer that has in extra
is
~Blue_Kite
Video Solution 1 by Math-X
https://youtu.be/Ku_c1YHnLt0?si=Xxe3CXtrXQYGSYik&t=3081
Video Solution 2 by CoolMathProblems
Video Solution 3 by hnsacademy
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=1453
Video Solution 4
~Education, the Study of Everything
Video Solution 5
Video Solution 6 by OmegaLearn (Using Cyclic Patterns)
Video Solution 7 by Magic Square
https://youtu.be/-N46BeEKaCQ?t=3990
Video Solution 8 by Interstigation
https://youtu.be/DBqko2xATxs&t=1845
Video Solution 9 by WhyMath
Video Solution 10
https://www.youtube.com/watch?v=jtHe6yLBz-4&t=20s
Video Solution 11 by Dr. David
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.