Difference between revisions of "2018 AMC 8 Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
The sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is <math>2^3=8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math> | The sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is <math>2^3=8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | There is also a pretty simple approach to this problem. Since in the bottom row you can either have 4 of the same signs, 3 of the same signs and one of another, and 2 of the same signs and one of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is <math>1 4 6 4 1</math>. Since 3 of one sign and 1 of the other doesn’t work, all you need to add is <math>1 + 6 + 1 = 8</math>, so the answer is <math>\boxed{\textbf{(C) } 8}</math> | ||
==See Also== | ==See Also== |
Revision as of 23:10, 9 November 2019
Problem 19
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
Instead of + and -, let us use 1 and 0, respectively. If we let ,
,
, and
be the values of the four cells on the bottom row, then the three cells on the next row are equal to
,
, and
taken modulo (mod) 2 (this is exactly the same as finding
, and so on). The two cells on the next row are
and
taken modulo (mod) 2, and lastly, the cell on the top row gets
.
Thus, we are looking for the number of assignments of 0's and 1's for ,
,
,
such that
, or in other words, is odd. As
, this is the same as finding the number of assignments such that
. Notice that, no matter what
,
, and
are, this uniquely determines
. There are
ways to assign 0's and 1's arbitrarily to
,
, and
, so the answer is
Solution 2
The sign of the next row on the pyramid depends on previous row. There are two options for the previous row, - or +. There are three rows to the pyramid that depend on what the top row is. Therefore, the ways you can make a + on the top is , so the answer is
Solution 3
There is also a pretty simple approach to this problem. Since in the bottom row you can either have 4 of the same signs, 3 of the same signs and one of another, and 2 of the same signs and one of the other, this can be thought of as the 4th Row of the Pascal’s Triangle, which is . Since 3 of one sign and 1 of the other doesn’t work, all you need to add is
, so the answer is
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 |
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