Difference between revisions of "2023 AMC 8 Problems/Problem 16"
MRENTHUSIASM (talk | contribs) (→Solution 1) |
MRENTHUSIASM (talk | contribs) (→Solution 3) |
||
Line 41: | Line 41: | ||
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</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>. | ||
~mathboy100 | ~mathboy100 | ||
− | |||
− | |||
==Animated Video Solution== | ==Animated Video Solution== |
Revision as of 15:51, 26 January 2023
Contents
Problem
The letters P, Q, and R are entered into a table according to the pattern shown below. How many Ps, Qs, and Rs will appear in the completed table?
Solution 1
In our grid we can see there are
,
and
of the letters P, Q and R’s respectively. We can see our pattern between each is
,
,
for the P, Q and 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
(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 by ILoveMath31415926535
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
,
, and
, and the second row has
,
, and
. Adding these up, we obtain
.
~mathboy100
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.