Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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So the answer is <math>\boxed{\textbf{(C) } 8}</math>. | So the answer is <math>\boxed{\textbf{(C) } 8}</math>. | ||
+ | |||
+ | Note: This result can also be achieved by realizing that there are <math>4! / 2! 2! = 6</math> ways to arrange <math>2</math> negatives and <math>2</math> positives and <math>1</math> way each to arrange four of one sign. | ||
~atharvd | ~atharvd | ||
+ | |||
+ | ~cxsmi (Note) | ||
+ | |||
+ | ==Solution 4== | ||
+ | The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a <math>+</math> on the above, half of them have a <math>-</math> above. So, For the lowest layer with <math>4</math> blocks, there are <math>2^4=16</math> possible combination and half of them will lead a <math>+</math> (or <math>-</math>) on the top. The answer is <math>16/2=\boxed{\textbf{(C) } 8}</math>. | ||
+ | |||
+ | If you notice this rule, you can give the answer whatever how many layers you have. The answer will be <math>2^{n-1}</math> for the layer with <math>n</math> blocks. | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We can use casework to solve this problem. | ||
+ | The only way for the top cell to have a <math>+</math> in it is if the third row of the pyramid (the one with <math>2</math> cells) is either <math>--</math> or <math>++</math>. First, let's pretend that the third row of the pyramid is <math>++</math>. The only way for that to happen is if the second row (the one with <math>3</math> cells) is --- or <math>+++</math>. Now, let's pretend that the second is <math>+++</math>. That would have <math>2</math> possibilities for the first row (the one with <math>4</math> cells), <math>++++</math> and <math>----</math>. Next, let's pretend that the second row is ---. That makes two more possibilities for the first row, <math>-+-+</math> and <math>+-+-</math>. Now, let's pretend that the 3rd row is <math>--</math>, which means that the second row is either <math>-+-</math> or <math>+-+</math>. You will soon find that <math>-+-</math> find <math>2</math> possibilities for the first row, <math>-++-</math> or <math>-++-</math>, and <math>2</math> possibilities for <math>+-+</math>, <math>--++</math> and <math>++--</math>. | ||
+ | Together, we find that the answer is <math>2+2+2+2=\boxed{\textbf{(C) } 8}</math>$ | ||
==Video Solution (CREATIVE ANALYSIS!!!)== | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
Line 87: | Line 102: | ||
~savannahsolver | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s | ||
==See Also== | ==See Also== |
Latest revision as of 14:57, 18 June 2024
Contents
Problem
In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
Solution 1
You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is .
-NinjaBoi2000
Solution 2
The top box is fixed by the problem.
Choose the left 3 bottom-row boxes freely. There are ways.
Then the left 2 boxes on the row above are determined.
Then the left 1 box on the row above that is determined
Then the right 1 box on that row is determined.
Then the right 1 box on the row below is determined.
Then the right 1 box on the bottom row is determined, completing the diagram.
So the answer is .
~BraveCobra22aops
Solution 3
Let the plus sign represent 1 and the negative sign represent -1.
The four numbers on the bottom are ,
,
, and
, which are either 1 or -1.
Which means = 1. Since
and
are either 1 or -1,
and
. This shows that
= 1.
Therefore either ,
,
, and
are all positive or negative, or 2 are positive and 2 are negative.
There are 2 ways where ,
,
, and
are 1 (1, 1, 1, 1) and (-1, -1, -1, -1)
There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1).
So the answer is .
Note: This result can also be achieved by realizing that there are ways to arrange
negatives and
positives and
way each to arrange four of one sign.
~atharvd
~cxsmi (Note)
Solution 4
The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a on the above, half of them have a
above. So, For the lowest layer with
blocks, there are
possible combination and half of them will lead a
(or
) on the top. The answer is
.
If you notice this rule, you can give the answer whatever how many layers you have. The answer will be for the layer with
blocks.
Solution 5
We can use casework to solve this problem.
The only way for the top cell to have a in it is if the third row of the pyramid (the one with
cells) is either
or
. First, let's pretend that the third row of the pyramid is
. The only way for that to happen is if the second row (the one with
cells) is --- or
. Now, let's pretend that the second is
. That would have
possibilities for the first row (the one with
cells),
and
. Next, let's pretend that the second row is ---. That makes two more possibilities for the first row,
and
. Now, let's pretend that the 3rd row is
, which means that the second row is either
or
. You will soon find that
find
possibilities for the first row,
or
, and
possibilities for
,
and
.
Together, we find that the answer is
$
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.